Konrad Schmüdgen The Moment Problemgolub/ref2/The-Moment-Problem.pdf · Konrad Schmüdgen Mathematisches Institut Universität Leipzig Germany ISSN 0072-5285 ISSN 2197-5612 (electronic)

Graduate Texts in Mathematics

Konrad Schmüdgen

The Moment Problem

Graduate Texts in Mathematics 277

Graduate Texts in Mathematics

Series Editors:

Sheldon AxlerSan Francisco State University, San Francisco, CA, USA

Kenneth RibetUniversity of California, Berkeley, CA, USA

Advisory Board:

Alejandro Adem, University of British ColumbiaDavid Eisenbud, University of California, Berkeley & MSRIIrene M. Gamba, The University of Texas at AustinJ.F. Jardine, University of Western OntarioJeffrey C. Lagarias, University of MichiganKen Ono, Emory UniversityJeremy Quastel, University of TorontoFadil Santosa, University of MinnesotaBarry Simon, California Institute of Technology

Graduate Texts in Mathematics bridge the gap between passive study andcreative understanding, offering graduate-level introductions to advanced topicsin mathematics. The volumes are carefully written as teaching aids and highlightcharacteristic features of the theory. Although these books are frequently used astextbooks in graduate courses, they are also suitable for individual study.

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http://www.springer.com/series/136

Konrad Schmüdgen

The Moment Problem

123

Konrad SchmüdgenMathematisches InstitutUniversität LeipzigGermany

ISSN 0072-5285 ISSN 2197-5612 (electronic)Graduate Texts in MathematicsISBN 978-3-319-64545-2 ISBN 978-3-319-64546-9 (eBook)DOI 10.1007/978-3-319-64546-9

Library of Congress Control Number: 2017951141

Mathematics Subject Classification (2010): 44A60, 14P10, 47A57

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This Springer imprint is published by Springer NatureThe registered company is Springer International Publishing AGThe registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To KATJA

Thomas Jean Stieltjes (1856 – 1894)

Pafnuty Lvovich Chebyshev Andrey Andreyevich Markov(1821 – 1894) (1856 – 1922)

Contents

1 Integral Representations of Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . 131.1 Integral Representations of Functionals on Adapted Spaces . . . . . 14

1.1.1 Moment Functionals and Adapted Spaces. . . . . . . . . . . . . . . . . 141.1.2 Existence of Integral Representations . . . . . . . . . . . . . . . . . . . . . 151.1.3 The Set of Representing Measures. . . . . . . . . . . . . . . . . . . . . . . . . 19

1.2 Integral Representations of Functionals onFinite-Dimensional Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.2.1 Atomic Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.2.2 Strictly Positive Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . 251.2.3 Sets of Atoms and Determinate Moment Functionals . . . . 271.2.4 Supporting Hyperplanes of the Cone of Moment

Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.2.5 The Set of Atoms and the Core Variety . . . . . . . . . . . . . . . . . . . 341.2.6 Extremal Values of Integrals with Moment

Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401.4 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2 Moment Problems on Abelian �-Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.1 �-Algebras and �-Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.2 Commutative �-Algebras and Abelian �-Semigroups . . . . . . . . . . . . 462.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.3.1 Example 1: Nd0; n

� D n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.3.2 Example 2: N2d

0 ; .n;m/� D .m; n/ . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.3.3 Example 3: Zd; n� D �n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.3.4 Example 4: Z; n� D n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532.5 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

vii

viii Contents

Part I The One-Dimensional Moment Problem

3 One-Dimensional Moment Problems on Intervals: Existence . . . . . . . . . 573.1 Positive Polynomials on Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.2 The Moment Problem on Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.3 The Symmetric Hamburger Moment Problem and Stieltjes

Moment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.4 Positive Polynomials on Intervals (Revisited) . . . . . . . . . . . . . . . . . . . . . 693.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.6 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4 One-Dimensional Moment Problems: Determinacy . . . . . . . . . . . . . . . . . . . . 794.1 Measures Supported on Bounded Intervals . . . . . . . . . . . . . . . . . . . . . . . . 794.2 Carleman’s Condition.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.3 Krein’s Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904.5 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

5 Orthogonal Polynomials and Jacobi Operators . . . . . . . . . . . . . . . . . . . . . . . . . 935.1 Definitions of Orthogonal Polynomials and Explicit Formulas. . . 945.2 Three Term Recurrence Relations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 985.3 The Moment Problem and Jacobi Operators . . . . . . . . . . . . . . . . . . . . . . 1025.4 Polynomials of the Second Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.5 The Wronskian and Some Useful Identities . . . . . . . . . . . . . . . . . . . . . . . 1085.6 Zeros of Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1125.7 Symmetric Moment Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1155.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1175.9 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

6 The Operator-Theoretic Approach to the Hamburger MomentProblem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.1 Existence of Solutions of the Hamburger Moment Problem.. . . . . 1216.2 The Adjoint of the Jacobi Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1236.3 Determinacy of the Hamburger Moment Problem . . . . . . . . . . . . . . . . 1256.4 Determinacy Criteria Based on the Jacobi Operator . . . . . . . . . . . . . . 1296.5 Self-Adjoint Extensions of the Jacobi Operator .. . . . . . . . . . . . . . . . . . 1336.6 Markov’s Theorem .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1376.7 Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1406.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1436.9 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

7 The Indeterminate Hamburger Moment Problem . . . . . . . . . . . . . . . . . . . . . . 1457.1 The Nevanlinna Functions A.z/;B.z/;C.z/;D.z/ . . . . . . . . . . . . . . . . . 1457.2 Von Neumann Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1497.3 Weyl Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1517.4 Nevanlinna Parametrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1537.5 Maximal Point Masses. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

Contents ix

7.6 Nevanlinna–Pick Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1607.7 Solutions of Finite Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1657.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1737.9 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

8 The Operator-Theoretic Approach to the Stieltjes MomentProblem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1778.1 Preliminaries on Quadratic Forms on Hilbert Spaces . . . . . . . . . . . . . 1778.2 Existence of Solutions of the Stieltjes Moment Problem . . . . . . . . . 1798.3 Determinacy of the Stieltjes Moment Problem .. . . . . . . . . . . . . . . . . . . 1808.4 Friedrichs and Krein Approximants.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1848.5 Nevanlinna Parametrization for the Indeterminate Stieltjes

Moment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1938.6 Weyl Circles for the Indeterminate Stieltjes Moment Problem .. . 1968.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1998.8 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

Part II The One-Dimensional Truncated Moment Problem

9 The One-Dimensional Truncated Hamburger and StieltjesMoment Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2039.1 Quadrature Formulas and the Truncated Moment Problem

for Positive Definite 2n-Sequences.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2049.2 Hamburger’s Theorem and Markov’s Theorem Revisited . . . . . . . . 2099.3 The Reproducing Kernel and the Christoffel Function.. . . . . . . . . . . 2119.4 Positive Semidefinite 2n-Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2149.5 The Hankel Rank of a Positive Semidefinite 2n-Sequence . . . . . . . 2179.6 Truncated Hamburger and Stieltjes Moment Sequences .. . . . . . . . . 2219.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2289.8 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

10 The One-Dimensional Truncated Moment Problem on aBounded Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22910.1 Existence of a Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22910.2 The Moment Cone SmC1 and Its Boundary Points . . . . . . . . . . . . . . . . 23110.3 Interior Points of SmC1 and Interlacing Properties of Roots . . . . . . 23410.4 Principal Measures of Interior Points of SmC1 . . . . . . . . . . . . . . . . . . . . 23710.5 Maximal Masses and Canonical Measures . . . . . . . . . . . . . . . . . . . . . . . . 24110.6 A Parametrization of Canonical Measures. . . . . . . . . . . . . . . . . . . . . . . . . 24610.7 Orthogonal Polynomials and Maximal Masses. . . . . . . . . . . . . . . . . . . . 24810.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25410.9 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

11 The Moment Problem on the Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25711.1 The Fejér–Riesz Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25711.2 Trigonometric Moment Problem: Existence of a Solution .. . . . . . . 25911.3 Orthogonal Polynomials on the Unit Circle . . . . . . . . . . . . . . . . . . . . . . . 261

x Contents

11.4 The Truncated Trigonometric Moment Problem . . . . . . . . . . . . . . . . . . 26611.5 Carathéodory Functions, the Schur Algorithm,

and Geromimus’ Theorem.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27111.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27811.7 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

Part III The Multidimensional Moment Problem

12 The Moment Problem on Compact Semi-Algebraic Sets . . . . . . . . . . . . . . 28312.1 Semi-Algebraic Sets and Positivstellensätze . . . . . . . . . . . . . . . . . . . . . . 28412.2 Localizing Functionals and Supports of Representing

Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28812.3 The Moment Problem on Compact Semi-Algebraic Sets

and the Strict Positivstellensatz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29312.4 The Representation Theorem for Archimedean Modules . . . . . . . . . 29812.5 The Operator-Theoretic Approach to the Moment Problem .. . . . . 30212.6 The Moment Problem for Semi-Algebraic Sets Contained

in Compact Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30712.7 Examples and Applications.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30812.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31112.9 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

13 The Moment Problem on Closed Semi-Algebraic Sets: Existence . . . . 31513.1 Positive Polynomials and Sums of Squares. . . . . . . . . . . . . . . . . . . . . . . . 31613.2 Properties (MP) and (SMP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32013.3 The Fibre Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32213.4 (SMP) for Basic Closed Semi-Algebraic Subsets

of the Real Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32613.5 Application of the Fibre Theorem: Cylinder Sets

with Compact Base . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33013.6 Application of the Fibre Theorem: The Rational Moment

Problem on Rd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33213.7 Application of the Fibre Theorem: A Characterization

of Moment Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33513.8 Closedness and Stability of Quadratic Modules . . . . . . . . . . . . . . . . . . . 33813.9 The Moment Problem on Some Cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34413.10 Proofs of the Main Implications of Theorems 13.10

and 13.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34813.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35313.12 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355

14 The Multidimensional Moment Problem: Determinacy . . . . . . . . . . . . . . . 35714.1 Various Notions of Determinacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35714.2 Polynomial Approximation.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36114.3 Partial Determinacy and Moment Functionals . . . . . . . . . . . . . . . . . . . . 36414.4 The Multivariate Carleman Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368

Contents xi

14.5 Moments of Gaussian Measure and Surface Measureon the Unit Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373

14.6 Disintegration Techniques and Determinacy . . . . . . . . . . . . . . . . . . . . . . 37414.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37814.8 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

15 The Complex Moment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38115.1 Relations Between Complex and Real Moment Problems.. . . . . . . 38115.2 The Moment Problems for the �-Semigroups

Zd and N0 �Zd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38415.3 The Operator-Theoretic Approach to the Complex

Moment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38615.4 The Complex Carleman Condition .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39115.5 An Extension Theorem for the Complex Moment Problem . . . . . . 39315.6 The Two-Sided Complex Moment Problem .. . . . . . . . . . . . . . . . . . . . . . 39515.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39715.8 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398

16 Semidefinite Programming and Polynomial Optimization. . . . . . . . . . . . . 39916.1 Semidefinite Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39916.2 Lasserre Relaxations of Polynomial Optimization with

Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40316.3 Polynomial Optimization with Constraints . . . . . . . . . . . . . . . . . . . . . . . . 40516.4 Global Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40816.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41016.6 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411

Part IV The Multidimensional Truncated Moment Problem

17 Multidimensional Truncated Moment Problems: Existence . . . . . . . . . . . 41517.1 The Truncated K-Moment Problem and Existence . . . . . . . . . . . . . . . 41617.2 The Truncated Moment Problem on Projective Space . . . . . . . . . . . . 42017.3 Hankel Matrices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42517.4 Hankel Matrices of Functionals with Finitely Atomic

Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42917.5 The Full Moment Problem with Finite Rank Hankel Matrix . . . . . 43317.6 Flat Extensions and the Flat Extension Theorem.. . . . . . . . . . . . . . . . . 43417.7 Proof of Theorem 17.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43817.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44117.9 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442

18 Multidimensional Truncated Moment Problems: BasicConcepts and Special Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44518.1 The Cone of Truncated Moment Functionals. . . . . . . . . . . . . . . . . . . . . . 44518.2 Inner Moment Functionals and Boundary

Moment Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45118.3 The Core Variety of a Linear Functional. . . . . . . . . . . . . . . . . . . . . . . . . . . 453

xii Contents

18.4 Maximal Masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45618.5 Constructing Ordered Maximal Mass Measures . . . . . . . . . . . . . . . . . . 46218.6 Evaluation Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46418.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46918.8 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470

19 The Truncated Moment Problem for Homogeneous Polynomials . . . . 47119.1 The Apolar Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47119.2 The Apolar Scalar Product and Differential Operators. . . . . . . . . . . . 47519.3 The Apolar Scalar Product and the Truncated

Moment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47719.4 Robinson’s Polynomial and Some Examples . . . . . . . . . . . . . . . . . . . . . . 48319.5 Zeros of Positive Homogeneous Polynomials .. . . . . . . . . . . . . . . . . . . . 48719.6 Applications to the Truncated Moment Problem on R2 . . . . . . . . . . 49119.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49619.8 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497

Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499A.1 Measure Theory .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499A.2 Pick Functions and Stieltjes Transforms .. . . . . . . . . . . . . . . . . . . . . . . . . . 502A.3 Positive Semidefinite and Positive Definite Matrices . . . . . . . . . . . . . 504A.4 Positive Semidefinite Block Matrices and Flat Extensions . . . . . . . 506A.5 Locally Convex Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509A.6 Convex Sets and Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510A.7 Symmetric and Self-Adjoint Operators on Hilbert Space . . . . . . . . . 514

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517

Symbol Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531

Preface and Overview

Let� be a positive Radon measure on a closed subset K of Rd and ˛ D .˛1; : : : ; ˛d/a multi-index of nonnegative integers ˛j. If the integral

s˛.�/ WDZKx˛11 : : : x

˛dd d�.x/

is finite, the number s˛ � s˛.�/ is called the ˛-th moment of the measure �. If all˛-th moments exist, the sequence .s˛/˛2Nd

0is called the moment sequence of �.

Moments appear at many places in physics and mathematics. For instance, if Kis a solid body in R3 with mass density m.x1; x2; x3/, then the number

s.0;2;0/ C s.0;0;2/ DZK.x22 C x23/m.x1; x2; x3/dx1dx2dx3

is the moment of interia of the body with respect to the x1-axis. Or if X is a randomvariable with distribution function F.x/, the expectation value of Xk is defined by

EŒXk� D sk DZ C1

�1xkdF.x/

and the variance of X is Var.X/ D EŒ.X � EŒX�/2� D EŒX2� � EŒX�2 D s2 � s21(provided that these numbers are finite).

The moment problem is a classical mathematical problem. In its simplest form,the Hamburger moment problem for the real line, it is the following question:

Let s D .sn/n2N0 be a real sequence. Does there exist a positive Radon measure� on R such that for all n 2 N0 the integral

R C1�1 xnd� converges and satisfies

sn DZ C1

�1xn d� ‹ (1)

1

2 Preface and Overview

That is, the moment problem is the inverse problem of “finding” a representingmeasure � when the moment sequence s is given.

For a real sequence s D .sn/n2N0 let Ls denote the linear functional on thepolynomial algebra RŒx� (or on CŒx�) defined by Ls.xn/ D sn, n 2 N0. By thelinearity of the integral it is clear that (1) holds for all n 2 N0 if and only if we have

Ls.p/ DZ C1

�1p.x/ d� for p 2 RŒx�: (2)

Thus, the moment problem asks whether or not the functional Ls on RŒx� admitsan integral representation (2) with respect to some positive measure �. To getsome flavour of what this book is about, we sketch without giving proofs somecornerstones from the theory of the one-dimensional Hamburger problem.

Assume that s D .sn/n2N0 is the moment sequence of some positive measure �on R. Then, for any polynomial p.x/ DPn

kD0 akxk 2 RŒx� we obtain

Ls.p2/ D

Zp.x/2 d� D

Z � nXk;lD0

akalxkCl

�d� D

nXk;lD0

akalskCl � 0:

Therefore, Ls.p2/ � 0 for all p 2 RŒx�, that is, the functional Ls is positive, and theHankel matrix Hn.s/ WD .skCl/

nk;lD0 is positive semidefinite for each n 2 N0. These

are two (equivalent) necessary conditions for a sequence to be a moment sequence.Hamburger’s theorem (1920) says that each of these necessary conditions is alsosufficient for the existence of a positive measure. That is, the existence problem fora solution is easily answered in terms of positivity conditions.

The question concerning the uniqueness of representing measures is more subtle.A moment sequence is called determinate if it has only one representing measure.For instance, the lognormal distribution

F.x/ D 1p2��.0;C1/.x/x

�1e�.log x/2=2

gives a probability measure whose moment sequence is not determinate.Let us assume that s has a representing measure � supported on the bounded

interval Œ�a; a�, a > 0. It is not difficult to show that the support of any otherrepresenting measure, say Q�, is also contained in Œ�a; a�. Then (2) implies that

Z a

�af .x/d� D

Z a

�af .x/ d Q� (3)

for f 2 RŒx�. By Weierstrass’ theorem each continuous functions on Œ�a; a� canbe approximated uniformly by polynomials. Therefore (3) holds for all continuousfunctions f on Œ�a; a�. Hence � D Q�, so that s is determinate.

Preface and Overview 3

A useful sufficient criterion for determinacy is the Carleman condition

1XnD1

s� 12n

2n D C1:

Now suppose s is a moment sequence such that Ls.pp/ > 0 for p 2 CŒx�; p ¤ 0,or equivalently, s has a representing measure with infinite support. Then

hp; qis D Ls.pq/; p; q 2 CŒx�;

defines a scalar product on the vector space CŒx�. Let X denote the multiplicationoperator by the variable x on CŒx�, that is, .Xp/.x/ D xp.x/. Applying the Gram–Schmidt procedure to the sequence .xn/n2N0 of the unitary space .CŒx�; h�; �is/ yieldsan orthonormal sequence .pn/n2N0 of polynomials pn 2 CŒx� such that each pn hasdegree n and a positive leading coefficient. Then there exist numbers an > 0 andbn 2 R such that the operator X acts on the orthonormal base .pn/ by

Xpn.x/ D anpnC1.x/C bnpn.x/C an�1pn�1.x/; n 2 N0; where p�1 WD 0:

That is, X is unitarily equivalent to the Jacobi operator TJ for the Jacobi matrix

J D

0BBBBB@

b0 a0 0 0 : : :

a0 b1 a1 0 : : :

0 a1 b2 a2 : : :

0 0 a2 b3 : : :

: : :: : :

: : :: : :

1CCCCCA

If A is a self-adjoint extension of TJ on a possibly larger Hilbert space and EA isthe spectral measure of A, then � D s0hEA.�/1; 1i is a solution of the momentproblem for s. Each solution is of this form. Further, the moment sequence s isdeterminate if and only if the closure of the symmetric operator X on the Hilbertspace completion of .CŒx�; h�; �is/ is self-adjoint. All these facts relate the momentproblem to the theory of orthogonal polynomials and to operator theory.

Let s be an indeterminate moment sequence. Nevanlinna’s theorem yields aparametrization of the solution set. Let P denote the set of holomorphic functionson the upper half-plane with nonnegative imaginary part and set P WD P [ f1g.Then Nevanlinna’s theorem states that there is a bijection � ! �� of P to the setof all solutions of the moment problems for s given by

Z 1

�11

x � zd��.x/ D �A.z/C �.z/C.z/

B.z/C �.z/D.z/ ; z 2 C; Im z > 0:


Here A;B;C;D are certain entire functions depending only on the sequence s. Thus,a moment sequence is either determinate or it has a “huge” family of representingmeasures. All this and much more is developed in detail in Part I of the book.

There are a number of other variants of the moment problem. The Stieltjesmoment problem asks for representing measures on the half-line Œ0;C1/ and theHausdorff moment problem for measures on the interval Œ0; 1�. If only the firstmoments s0; : : : ; sm are prescribed, we have a truncated moment problem. In thiscase there exist finitely atomic representing measures and one is interested inmeasures with “small” numbers of atoms.

The passage from one-dimensional to multidimensional moment problems leadsto fundamental new difficulties. As already observed by Hilbert, there are positivepolynomials in d � 2 variables that are not sums of squares. As a consequence,there exists a linear functional L on RdŒx� � RŒx1; : : : ; xd� which is positive (thatis, L.p2/ � 0 for p 2 RdŒx�), but L is not a moment functional (that is, it cannot bewritten in the form (2)). For this reason the K-moment problem is invented. It asksfor representing measures with support contained in a given closed subset K of Rd.

Let ff1; : : : ; fkg be a finite subset of RdŒx�. Then the closed set

K D fx 2 Rd W f1.x/ � 0; : : : ; fk.x/ � 0g

is called a semi-algebraic set. For such sets methods from real algebraic geometryprovide powerful tools for the study of the K-moment problem. Our main result fora compact semi-algebraic set K is the following: A linear functional L on RdŒx� is amoment functional with representing measure supported on K if and only if

L.f e11 � � � f ekk p2/ � 0 for p 2 RdŒx�; e1; : : : ; ek 2 f0; 1g:

If K is only closed but not compact, this is no longer true. But if there exist nontrivialbounded polynomials on K, there is a fibre theorem that allows one to reduce the K-moment problem to “lower dimensional” cases. The multidimensional determinacyquestion is much more complicated than its one-dimensional counterpart.

It turns out that most methods that have been successfully applied to the momentproblem in dimension one either fail in higher dimensions or at least require moreinvolved additional technical considerations.

The study of moment problems and related topics goes back to the late nineteenthcentury. The Russian mathematicians P.L. Chebychev (1974) and A.A. Markov(1984) applied moments in their “theory of limiting values of integrals” and inventedimportant notions; a survey of their ideas and further developments was givenby M.G. Krein [Kr2]. The moment problem itself as a problem in its own wasformulated for the first time by the Dutch mathematician T.J. Stieltjes (1894) inhis pioneering memoir [Stj]. Stieltjes treated this problem for measures supportedon the half-line and developed a far reaching theory. The cases of the real line and ofbounded intervals were studied only later by H. Hamburger (1920) and F. Hausdorff


(1920). Important early contributions have been made by R. Nevanlinna, M. Riesz,T. Carleman, M.H. Stone, and others.

A surprising feature of the moment problem theory is the connections andthe close interplay with many branches of mathematics and the broad range ofapplications. H.J. Landau wrote in the introduction of an article in the AMS volume“Moments in Mathematics” [L2]: “The moment problem is a classical question inanalysis, remarkable not only for its own elegance, but also for its extraordinaryrange of subjects theoretical and applied, which it has illuminated. From it flowdevelopments in function theory, in functional analysis, in spectral representationof operators, in probability and statistics, in Fourier analysis and the prediction ofstochastic processes, in approximation and numerical methods, in inverse problemsand the design of algorithms for simulating physical systems.” Looking at thedevelopments of the multidimensional moment problem over the last two decadesI would like to add real algebraic geometry, optimization, and convex analysis toLandau’s list.

This book is an advanced text on the moment problem on Rd and its moderntechniques. It is divided into four main parts, two devoted to one-dimensionalmoment problems and two others to multidimensional moment problems. In eachgroup we distinguish between full and truncated moment problems. Though ourmain emphasis is on real moment problems we include short treatments of themoment problem on the unit circle and of the complex moment problem.

Here is a brief description of the four parts.Part I deals with the one-dimensional full moment problem and develops

important methods and technical tools such as orthogonal polynomials, Jacobioperators, and Nevanlinna functions in great detail. Basic existence and uniquenesscriteria are obtained, but also a number of advanced results such as the Nevanlinnaparametrizations for indeterminate Hamburger and Stieltjes problems, finite ordersolutions, Nevanlinna–Pick interpolation, Krein and Friedrichs approximants ofStieltjes solutions, and others are included.

Part II is about one-dimensional truncated moment problems. The truncatedHamburger and Stieltjes moment problems are treated and Gauss’ quadratureformulas are derived. In the case of bounded intervals the classical theory ofMarkoff, Krein, and Akhiezer on canonical and principal measures, maximalmasses, and the moment cone are studied. Part II also contains a self-containeddigression to the trigonometric moment problem and some highlights of this theory(Schur algorithm, Verblunsky and Geronimus’ theorems).

In Part III the multidimensional full moment problem on closed semi-algebraicsubsets of Rd is investigated. Here real algebraic geometry and operator theoryon Hilbert space are the main tools. For compact semi-algebraic sets the interplaybetween strict Positivstellensätze and the moment problem leads to satisfactoryexistence results for the moment problem. In the case of closed semi-algebraic setsexistence criteria and determinacy are much more subtle. The fibre theorem andits applications and the multidimensional determinacy problem are investigated indetail. Further, we derive basic existence and determinacy results for the complex


moment problem. The two main Chaps. 12 and 13 are the heart of Part III andalso of the book. At the end of Part III we touch very briefly upon semidefiniteprogramming and applications of moment methods to polynomial optimization.

Part IV gives an introduction to the multidimensional truncated moment problem.Existence theorems in terms of positivity and the flat extension theorem are derived.Fundamental technical tools (Hankel matrices, evaluation polynomials, the apolarscalar product for homogeneous polynomials) are developed and discussed in detail.A number of important special topics (the core variety, maximal masses, deter-minacy, Carathéodory numbers) are also studied. The multidimensional truncatedmoment problem is an active topic of present research. It is expected that convexanalysis and algebraic geometry will provide new powerful methods and that thestatus of this area might essentially change in the coming decades. For this reason,we have not treated all recent developments; instead we have concentrated on basicresults and concepts and on selected special topics.

All moment problems treated in this book deal with integral representations oflinear functionals on a commutative unital algebra or on a (in most cases finite-dimensional) vector space of continuous functions on a locally compact space. Mostof them are moment problems on certain �-semigroups. In two introductionarychapters, general results on integral representations of positive functionals areobtained and notions concerning moment problems for �-semigroups are developed.These results and notions will play an essential role throughout the whole book.

As mentioned above, the main focus of this book is on the four versions of thescalar classical moment problem. In the course of this we develop fundamentalconcepts and technical tools and we derive deep classical theorems and very recentresults as well. Also, we present a number of new results and new proofs. In thisbook, we do not treat matrix moment problems, operator moment problems, infinite-dimensional moment problems, or noncommutative moment problems.

Apart from the two introductory chapters the parts of the book and a number ofchapters can be read (almost) independently from each other. Sometimes a technicalfact from another part is used in a proof; it can be filled easily. In order to beindependent from previous chapters we have occasionally repeated some notation.

Several courses and seminars on moment problems can be built on this book bychoosing appropriate material from various parts. Each course or seminar shouldprobably start with the corresponding results from Sects. 1.1 and 1.2. For a onesemester basic course on the one-dimensional moment problem this could befollowed by Sects. 2.1–2.2, Chap. 3, and the core material of the first sections ofChaps. 4–6. A one semester advanced course on the multidimensional momentproblem could be based on Sects. 11.1–11.3, 11.5–11.6, and selected material fromChaps. 12 and 13, avoiding technical subtleties. Here applications to optimizationfrom Chap. 15 are optional. Chapters 8, 9, 10, and 16 are almost self-contained andcould be used for special seminars on these topics. Most exercises at the end ofeach chapter should be solvable by active students; some of them contain additionalresults or information on the corresponding topics.

As the title of the book indicates, the real moment problem is the central topic.Our main emphasis is on a rigorous treatment of the moment problem, but also


on important methods and technical tools. Often several proofs or approaches tofundamental results are given. For instance, Hamburger’s theorem 3.8 is derivedin Sect. 3.2 from Haviland’s theorem, in Sect. 6.1 from the spectral theorem, and inSect. 9.2 from the truncated moment problem. This is also true for a number of otherresults such as Stieltjes’ theorem, Carleman’s theorem, Markov’s theorem, and thePositivstellensätze and moment problem theorems in Chap. 12. At various placesexplicit formulas in terms of the moments are provided even if they are not neededfor treating the moment problem. Large parts of the material, especially in Parts IIIand IV, appear for the first time in a book and a number of results are new.

Necessary prerequisities for this book are a good working knowledge of measureand integration theory and of polynomials, but also the basics of holomorphicfunctions in one variable, functional analysis, convex sets, and elementary topology.With this background about two thirds of the book should be readable by graduatestudents. In the remaining third (more precisely, in parts of Chaps. 4–7 and Part III)Hilbert space operator theory and real algebraic geometry play an essential role.The short disgression into real algebraic geometry given in Sect. 12.1 covers allthat is needed; for more details we refer to the standard books [Ms1] and [PD].Elementary facts on unbounded symmetric or self-adjoint operators and the spectraltheorem (multidimensional versions in Sect. 12.5 and 15.3) are used at variousplaces; necessary notions and facts are collected in Appendix A.7. In Chap. 8Friedrichs and Krein extensions of positive symmetric operators occur; they arebriefly explained in Sect. 8.1. All operator-theoretic notions and results needed inthis book can be found (for instance) in the author’s Graduate Text [Sm9].

In large parts of the book results and techniques from other mathematicalfields are used. In most cases we state or develop such results with referenceto the literature at the places where they are needed. Further, I have added sixappendices: on measure theory, on Pick functions and Stieltjes transforms, onpositive semidefinite matrices, on locally convex topologies, on convex sets, andon Hilbert space operators. These appendices collect notions and facts that are usedoften and at different places of the text. For some results we have included proofs.

Some general notation is collected after the table of contents. Though I tried toretain standard terminology in most cases, occasionally I have made some changes,we hope for the better. For instance, instead of the term “moment matrix” I preferred“Hankel matrix” and denoted it by Hn.L/ or H.L/: Also there is some overlappingnotation. While the symbol A is used for one of the four Nevanlinna functions inChap. 7 (following standard notation), it denotes a matrix or a Hilbert space operatorat other places. The meaning will be always clear from the context. The underlyingalgebras are usually denoted by sanserif letter such as A, B:

Continued fractions are avoided in this book (in Sect. 6.7 only the notion isbriefly explained). Instead I have put my emphasis on operator-theoretic approaches,because I am convinced that these methods are more promising and powerful, inparticular concerning the multidimensional case.

In writing this book I benefited very much from N.I. Akhiezer’s classic [Ak]and B. Simon’s article [Sim1], but also from standard books such as [BCRl], [KN],[KSt], [Ms1], [Sim3], [Chi1] and from the surveys [La2], [AK]. Applications and


ramifications of moment problems and related topics are discussed and developedin the AMS volume [L2] and in the books [Ls2], [BW].

I feel unable to give precise credits for all results occuring in this book. In theNotes at the end of each chapter I have given (to the best of my knowledge) creditsfor some results, including the main theorems, and some hints to the literature andfor further reading. In the bibliography I have listed some key classical papers.

I am indebted to B. Reznick, C. Scheiderer, and J. Stochel for valuable commentson parts of the manuscripts. I am grateful to Ph. di Dio for reading the whole textand for his helpful suggestions. Also, I should like to thank R. Lodh and A.-K.Birchley-Brun from Springer-Verlag for their help in publishing this book.

Leipzig, Germany Konrad SchmüdgenMay 7, 2017

General Notations

Numbers

N0 nonnegative integersN positive integersZ integersR real numbersC complex numbersT complex numbers of modulus oneD complex numbers of modulus less than oneRC D Œ0;C1/CC D fz 2 C W Im z > 0gi complex unit

Spaces and Sets

Rd d-dimensional real spaceCd d-dimensional complex spacePd.R/ d-dimensional real projective spaceTd d-torusSd�1 unit sphere in Rd

Matrices

Mn;k.K/ .n; k/-matrices over K, Mn.K/ D Mn;n.K/

Symn symmetric matrices in Mn.R/

AT transposed matrix of A

Measures

MC.X / (positive) Radon measures on a locally compact Hausdorff space XM.X / complex Radon measures on XL1.X ; �/ �-integrable Borel functions on Xıx delta measure at the point x�M characteristic function of a set MRadon measures are always nonnegative!

9

10 General Notations

Polynomials

x˛ D x˛11 � � � x˛dd for ˛ D .˛1; � � � ; ˛d/ 2 Nd0

RdŒx� D RŒx1; : : : ; xd�RdŒx�n D fp 2 RdŒx� W deg.p/ � ngHd;m homogeneous polynomials from RdŒx� of degree mqh homogenization of qPos.K/ D ff 2 E W f .x/ � 0; x 2 Kg, where E C.X IR/; K XPos.A;K/ D fp 2 A W p.x/ � 0; x 2 KgZ.p/ D fx 2 Rd W p.x/ D 0g for p 2 RdŒx�d is the number of variables and n;m denote the degrees of polynomials!

Moments, Moment Sequences, and Measures

s˛ DRx˛ d� ˛-the moment of �

s D .s˛/ moment sequences.x/; sN.x/ moment vector of the delta measure ıxSmC1;S;S.A;K/ moment conesMC.Rd/ D f� 2 MC.Rd/ W R jx˛j d� < C1 for ˛ 2 Nd

0g Radon measureswith finite moments

Ms D f� 2 MC.Rd/ W s˛ DRx˛ d� for ˛ 2 Nd

0g representing measures of sLs Riesz functional associated with s defined by Ls.x˛/ D s˛Hn.s/ finite Hankel matrix associated with sH.s/ infinite Hankel matrix associated with sDn.s/ D detHn.s/ Hankel determinant

Orthogonal Polynomials and Functions

pn orthonormal polynomial of the first kindPn monic orthogonal polynomial of the first kindqn orthogonal polynomial of the second kindQn monic orthogonal polynomial of the second kindpz D .p0.z/; p1.z/; p2.z/; : : : /qz D .q0.z/; q1.z/; q2.z/; : : : /P Pick functions�s Friedrichs parameterA.z/;B.z/;C.z/;D.z/ entire functions in the Nevanlinna parametrization

Operators

X multiplication operator by the variable xJ infinite Jacobi matrixT D TJ Jacobi operator for the Jacobi matrix Jd D f.c0; c1; : : : ; cn; 0; 0; : : : / W cj 2 C; n 2 N0g finite complex sequenceshp; qis D Ls.pq/ scalar product on CŒx� associated with sHs Hilbert space completion of .CŒx�; h�; �is/X1; : : : ;Xd multiplication operators by the variables x1; : : : ; xd

General Notations 11

hp; qiL D L.pq/ scalar product on CdŒx� associated with LHL Hilbert space completion of .CdŒx�; h�; �iL/�L GNS representation associated with LTF Friedrichs extension of a positive symmetric operator TTK Krein extension of a positive symmetric operator T�.T/ spectrum of T.T/ resolvent set of TD.T/ domain of TN .T/ null space of TET spectral measure of T

Real Algebraic Geometry

T.f/ preordering generated by f, where f is a finite subset of RdŒx�Q.f/ quadratic module generated by fK.f/ semi-algebraic set defined by fRŒV� algebra of regular functions on a real algebraic set VP2

n sum of squares p2 of polynomials p 2 RdŒx�, wheredeg p � n

OA characters of a real unital algebra APA2 sum of squares a2 of elements a of a real algebra A

AC D ff 2 A W �.f / � 0; � 2 OA gMoment Functionals and Related Sets

L;L.A;K/ cones of moment functionalslx point evaluation functional at xLg D L.g�/ localization of L at gML D f� 2 MC.Rd/ W L.p/D R p.x/ d� for p2RdŒx�g representing measures of LL.t/ maximal mass of representing measures of L at tH.L/ Hankel matrix of LW.L/;W.L/ set of atoms of representing measures of LV.L/;V.L/ core variety of LNL D ff 2 A W L.fg/ D 0; g 2 A gVL D ft 2 Rd W f .t/ D 0; f 2 NL gNC.L;K/ D ff 2 Pos.A;K/ W L.f / D 0 gVC.L;K/ D ft 2 Rd W f .t/ D 0; f 2 NC.L;K/ gNC.L/ D NC.L;Rd/

VC.L/ D V.L;Rd/

NC.L/ D ff 2 EC W L.p/ D 0 gVC.L/ D fx 2 X W f .x/ D 0; f 2 NC.L/ gSets are denoted by braces such as fxi W i 2 Ig, while sequences are written as.xn/n2N or .xn/.

Chapter 1Integral Representations of Linear Functionals

All variants of moment problems treated in this book deal with following problem:Given a linear functional L on a vector space E of continuous functions on a

locally compact Hausdorff space X and a closed subset K of X , when does thereexist a (positive) Radon measure � supported on K such that

L. f / DZXf .x/ d�.x/ for f 2 E‹

Functionals of this form are called K-moment functionals or simply momentfunctionals when K D X . In this chapter, we develop the underlying basic setupand introduce a number of general notions.

In Sect. 1.1, we prove various integral representation theorems for functionals onadapted spaces (Theorems 1.8, 1.12, and 1.14) and derive properties of the set ofrepresenting measures (Theorems 1.19, 1.20, and 1.21). Our existence theorems forfull moment problems derived in Parts I and III are based on these results.

Section 1.2 is devoted to the case when E has finite dimension. Then, bythe Richter–Tchakaloff theorem (Theorem 1.24), each moment functional has afinitely atomic representing measure. Strictly positive linear functionals (The-orem 1.30), determinate moment functionals (Theorem 1.36) and the cone ofmoment functionals are investigated. Further, we study the set of possible atomsof representing measures (Theorem 1.45) and prove that it coincides with the corevariety (Theorem 1.49). The last subsection deals with extreme values of integralsRhd�, where the measure � has fixed moments (Theorems 1.50 and 1.52). The

results obtained in Sect. 1.2 are useful for truncated moment problems treated inParts II and IV, but most of them are also of interest in themselves.

Throughout this chapter, X denotes a locally compact topological Hausdorffspace, E is a linear subspace of the space C.X IR/ of real-valued continuousfunctions on X , and L is a linear functional on E.

© Springer International Publishing AG 2017K. Schmüdgen, The Moment Problem, Graduate Texts in Mathematics 277,DOI 10.1007/978-3-319-64546-9_1

13

14 1 Integral Representations of Linear Functionals

1.1 Integral Representations of Functionalson Adapted Spaces

Recall that MC.X / denotes the set of Radon measures on X and that in ourterminology Radon measures are always nonnegative (see Appendix A.1).

1.1.1 Moment Functionals and Adapted Spaces

If C is a subset of E, the functional L is called C-positive if L. f / � 0 for f 2 C. Set

EC WD f f 2 E W f .x/ � 0 for all x 2 X g:

If � is a measure from MC.X / such that E L1.X ; �/, we define an EC-positivelinear functional L� on E by

L�. f / DZX

f .x/d�.x/; f 2 E: (1.1)

The following two definitions introduce basic notions that will be used throughoutthis book. The terminology “moment functionals” will be clear later when we studymoment functionals on examples of �-semigroups, see Sect. 2.3.1.

Definition 1.1 A linear functional L on E is a moment functional if there exists ameasure� 2 MC.X / such that L D L�. Any such measure� is called a representingmeasure of L. The set of all representing measures of L is

ML D f� 2 MC.X / W L D L�g:

A moment functional L is called determinate if it has a unique representingmeasure, or equivalently, if the set ML is a singleton.

Definition 1.2 Let K be a closed subset of X : A functional L on E is a K-momentfunctional if there exists a measure � 2 MC.X / supported on K such that L D L�.The set of such measures is

ML;K D f� 2 MC.X / W L D L� and supp� K g:

A K-moment functional L is said to be K-determinate if the set ML;K is a singleton.

The aim of this section is to apply Choquet’s concept of adapted spaces to thestudy of moment functionals.

Definition 1.3 For f ; g 2 C.X IR/ we say that g dominates f if for any " > 0 thereexists a compact subset K" of X such that j f .x/j � "jg.x/j for all x 2 XnK".

Roughly speaking, g dominates f means that j f .x/=g.x/j ! 0 as x!1.

1.1 Integral Representations of Functionals on Adapted Spaces 15

We give a slight reformulation of the domination property and set

U WD f 2 Cc.X IR/ W 0 � .x/ � 1 for x 2 X g: (1.2)

Lemma 1.4 For any f ; g 2 C.X IR/ the following statements are equivalent:(i) g dominates f .

(ii) For " > 0 there is an " 2 U such that j f .x/j � "jg.x/j C j f .x/j".x/, x 2 X :(iii) For " > 0 there is an h" 2 Cc.X IR/ such that j f .x/j � "jg.x/jCh".x/, x 2 X :

Proof(i)!(ii) Choose " 2 U such that ".x/ D 1 on K":(ii)!(iii) is clear by setting h" WD j f j".(iii)!(i) Since h" 2 Cc.X IR/, the set K" WD supp h" is compact and we have

j f .x/j � "jg.x/j for x 2 XnK". utDefinition 1.5 A linear subspace E of C.X IR/ is called adapted if the followingconditions are satisfied:

(i) E D EC � EC.(ii) For each x 2 X there exists an f 2 EC such that f .x/ > 0.

(iii) For each f 2 EC there exists a g 2 EC such that g dominates f .

Lemma 1.6 If E is an adapted subspace of C.X IR/, then for any f 2 Cc.X IR/Cthere exists a g 2 EC such that g.x/ � f .x/ for all x 2 X .

Proof Let x 2 X . By Definition 1.5(ii) there exists a function gx 2 EC such thatgx.x/ > 0. Multiplying gx by some positive constant we get gx.x/ > f .x/. Thisinequality remains valid in some neighbourhood of x. By the compactness of supp fthere are finitely many x1; : : : ; xn 2 X such that g.x/ WD gx1 .x/C� � �Cgxn.x/ > f .x/for x 2 supp f and g.x/ � f .x/ for all x 2 X . ut

1.1.2 Existence of Integral Representations

In the proof of Theorem 1.8 below we use the following extension theorem.

Proposition 1.7 Let E be a linear subspace of a real vector space F and let C be aconvex cone of F such that F D ECC. Then each .C\E/-positive linear functionalL on E can be extended to a C-positive linear functional QL on F.

Proof Let f 2 F. We define

q. f / D inf fL.g/ W g 2 E; g � f 2 Cg: (1.3)

Since F D ECC, there exists a g 2 E such that�fCg 2 C, so the corresponding setin (1.3) is not empty. It is easily seen that q is a sublinear functional and L.g/ D q.g/


for g 2 E. Therefore, by the Hahn–Banach theorem, there is an extension QL of L toF such that QL. f / � q. f / for all f 2 F.

Let h 2 C. Setting g D 0; f D �h we have g� f 2 C, so that q.�h/ � L.0/ D 0by (1.3). Hence QL.�h/ � q.�h/ � 0, so that QL.h/ � 0. Thus, QL is C-positive. ut

Most existence results on the moment problem derived in this book have theirorigin in the following theorem.

Theorem 1.8 Suppose that E is an adapted subspace of C.X IR/. For any linearfunctional L W E! R the following are equivalent:

(i) The functional L is EC-positive, that is, L. f / � 0 for all f 2 EC.(ii) For each f 2 EC there exists an h 2 EC such that L. f C "h/ � 0 for all " > 0.

(iii) L is a moment functional, that is, there exists a measure � 2 MC.X / such thatL D L�.

Proof The implications (i)!(ii) and (iii)!(i) are clear.(ii)!(i) Let f 2 EC. Letting "! 0 in the inequality L. f /C"L.h/ D L. fC"h/ �

0, we get L. f / � 0.(i)!(iii) We begin by setting

QE WD f f 2 C.X ;R/ W j f .x/j � g.x/; x 2 X ; for some g 2 Eg

and claim that QE D E C . QE/C. Obviously, E C . QE/C QE. Conversely, let f 2 QE.We choose a g 2 EC such that j f j � g. Then we have f C g 2 . QE/C, �g 2 E and�f D �gC .gC f / 2 EC . QE/C. That is, QE D EC . QE/C.

By Proposition 1.7, L can be extended to an . QE/C-positive linear functional QL onQE. We have Cc.X IR/ QE by Lemma 1.6. From the Riesz representation theorem(Theorem A.4) it follows that there is a measure � 2 MC.X / such that QL. f / DRf d� for f 2 Cc.X IR/. By Definition 1.5(i), E D EC � EC. To complete the

proof it therefore suffices to show that f 2 L1.X ; �/ and L. f / � QL. f / D R f d� forf 2 EC.

Fix f 2 EC. Let U be the set defined by (1.2). For 2 U , f 2 Cc.X IR/ andhence QL. f/ D R f d�. Using this fact and the . QE/C-positivity of QL, we derive

Zfd� D sup

2U

Zf d� D sup

2UQL. f/ � QL. f / D L. f / <1; (1.4)

so that f 2 L1.X ; �/.By (1.4) it suffices to prove that L. f / � R

fd�. From Definition 1.5(iii), thereexists a g 2 EC that dominates f . Then, by Lemma 1.4, for any " > 0 there exists afunction " 2 U such that f � "gC f". Since f" � f , we obtain

L. f / D QL. f / � " QL.g/C QL. f"/ D "L.g/CZ

f"d� � "L.g/CZ

fd�:


Note that g does not depend on ". Passing to the limit "!C0, we get L. f / � R fd�.Thus, L. f /D R fd� which completes the proof of (iii). ut

If X is compact, then C.X IR/ D Cc.X IR/, so condition (iii) in Definition 1.5is always fulfilled and can be omitted. But in this case we can obtain the desiredintegral representation of L more directly, as the following proposition shows.

Proposition 1.9 Suppose that X is a compact Hausdorff space and E is a linearsubspace of C.X IR/ which contains a function e such that e.x/ > 0 for x 2 X .

Then each EC-positive linear functional L on E is a moment functional, that is,there exists a measure � 2 MC.X / such that L. f / D

Rf d� for f 2 E.

Proof Set F D C.X IR/ and C D C.X IR/C. Let f 2 F. Since X is compact,f is bounded and e has a positive infimum. Hence there exists a � > 0 such thatf .x/ � �e.x/ on X . Since �e � f 2 C and ��e 2 E, f D ��eC .�e � f / 2 EC C.Thus, F D EC C. By Proposition 1.7, L extends to a C-positive linear functional QLon F. By the Riesz representation theorem, QL, hence L, can be given by a measure� 2 MC.X /. utRemark 1.10 If X is compact, the assumption e.x/ > 0 on X in Proposition 1.9implies that e is an interior point of the cone EC. This is a standard assumption ofthe theory of ordered vector spaces which will be used in Theorem 1.26 below aswell. In applications in this book we usually take e D 1. ı

In the proof of Theorem 1.8 condition (iii) of Definition 1.5 was crucial. We givea simple example where this condition fails and L has no representing measure.

Example 1.11 Set E WD Cc.RIR/CR � 1 and define a linear functional on E by

L. f C � � 1/ WD � for f 2 Cc.RIR/; � 2 R;

where 1 denotes the constant function equal to 1 on R. Then L is EC-positive, but itis not a moment functional. (Indeed, since L. f / D 0 for f 2 Cc.RIR/, (1.1) wouldimply that the measure � is zero. But this is impossible, because L.1/ D 1.) ı

The next result is called Haviland’s theorem. For a closed subset K of Rd we set

Pos.K/ D f p 2 RdŒx� W p.x/ � 0 for all x 2 K g:

Theorem 1.12 Let K be a closed subset of Rd and L a linear functional on RdŒx�.The following statements are equivalent:

(i) L. f / � 0 for all f 2 Pos.K/.(ii) L. f C "1/ � 0 for f 2 Pos.K/ and " > 0.

(iii) For any f 2 Pos.K/ there is an h 2 Pos.K/ such that L. f C "h/ � 0 for all" > 0.

(iv) L is a K-moment functional, that is, there exists a measure � 2 MC.Rd/

supported on K such that f 2 L1.Rd; �/ and L. f / D RK f d� for all f 2 RdŒx�.


Proof Set X D K. Then E D RdŒx� is an adapted subspace of C.K;R/. Indeed,condition (i) in Definition 1.5 follows from the relation 4p D . pC 1/2 � . p � 1/2.Condition (ii) is trivial. If p 2 EC, then g D .x21 C � � � C x2d/f dominates f , socondition (iii) is also fulfilled.

Note that the implications (iv)!(i)!(ii)!(iii) in Theorem 1.12 are obvious. Theother assertions follow from Theorem 1.8. ut

Now suppose that A is a finitely generated commutative real unital algebra. Wedevelop some notation and facts that will be used in Chaps. 12 and 13.

Definition 1.13 A character of A is an algebra homomorphism � W A ! R

satisfying �.1/ D 1. The set of characters of A is denoted by OA.

We fix a set f f1; : : : ; fdg of generators of A. Then there exists a unique surjectiveunital algebra homomorphism � W RdŒx�! A such that �.xj/ D fj; j D 1; : : : ; d. IfJ denotes the kernel of � , then J is an ideal of RdŒx� and A is isomorphic to thequotient algebra RdŒx�=J ; that is,

A Š RdŒx�=J :

Each character � of A is uniquely determined by the point x� WD .�. f1/; : : : ; �. fd//of Rd. We identify � with x� and write f .x�/ WD �. f / for f 2 A. Under thisidentification, OA becomes the real algebraic set

OA D Z.J / WD fx 2 Rd W p.x/ D 0 for p 2 J g: (1.5)

Since Z.J / is closed in Rd, OA is a locally compact Hausdorff space in the inducedtopology of Rd and elements of A can be considered as continuous functions on OA:In the case A D RdŒx� we can take p1 D x1; : : : ; pd D xd and obtain OA D Rd.

For a closed subset K of OA, we define

Pos.K/ D f f 2 A W f .x/ � 0 for x 2 K g:

Then we have the following generalized version of Haviland’s theorem for A.

Theorem 1.14 Let A be a finitely generated commutative real unital algebra and Ka closed subset of OA. For a linear functional L on A, the following are equivalent:

(i) L. f / � 0 for all f 2 Pos.K/.(ii) L. f C "1/ � 0 for f 2 Pos.K/ and " > 0.

(iii) For any f 2 Pos.K/ there is an h 2 Pos.K/ such that L. f C "h/ � 0 for all" > 0.

(iv) L is a K-moment functional, that is, there exists a measure � 2 MC. OA/supported on K such that A L1. OA; �/ and L. f / D ROA f d� for all f 2 A.

Proof (iv)!(i)!(ii)!(iii) are obviously satisfied. (iii)!(i) follows as in the proofof Theorem 1.8. It remains to prove the main implication (i)!(iv).


We define a linear functional QL on RdŒx� by QL D L ı � . Let p 2 RdŒx� andset f WD �. p/. Since � is an algebra homomorphism and �.xj/ D fj, we havep. f1; : : : ; fd/ D p.�.x1/; : : : ; �.xd// D �. p.x// D f . Hence, for x 2 K,

f .x/D�x. f /D�x. p. f1; : : : ; fd//D p.�x. f1/; : : : ; �x. fd//D p.x1; : : : ; xd/D p.x/:

Thus, if p.x/ � 0 on K, then �. p/ D f � 0 on K and hence QL. p/ D L. f / � 0by assumption (i). Therefore, by Theorem 1.12 there exists a measure � 2 MC.Rd/

such that supp� K and QL. p/ D RK p d� for all p 2 RdŒx�.Let f 2 A. Then f D �. p/ for some p 2 RŒx�. Using the definition of QL and the

equality p.x/ D f .x/ for x 2 K we obtain

L. f / D L.�. p// D QL. p/ DZKp.x/ d�.x/ D

ZKf .x/ d�.x/:

This proves (iv). utRemark 1.15 Let A and K be as in Theorem 1.14. Let L.K/ denote the K-momentfunctionals on A. Then Pos.K/ and L.K/ are cones in A resp. in its dual satisfying

Pos.K/^ D L.K/ and L.K/^ D Pos.K/; (1.6)

where the dual cones Pos.K/^ and L.K/^ are defined by (A.19). Thus, the conesPos.K/ and L.K/ are dual to each other. Indeed, the first equality of (1.6) isTheorem 1.14(i)$(ii), while the second follows from the bipolar theorem [Cw,Theorem V.1.8]. ıExample 1.16 Let A be the unital real algebra of functions on R generated by thetwo functions f1 D 1

1Cx2and f2 D x

1Cx2. Then the identity . f1 � 1

2/2 C f 22 D 1

4holds

in A and the set OA is given by the points of the circle in R2 with center . 12; 0/ and

radius 12. Note that the character � 2 OA with �. f1/ D �. f2/ D 0 cannot be obtained

by a point evaluation on R. ı

1.1.3 The Set of Representing Measures

In this subsection we use some facts concerning the vague convergence of measures,see Appendix A.1.

Definition 1.17 A directed net .Li/i2I of linear functionals on E is a net of linearfunctionals Li defined on vector subspaces Ei, i 2 I, of E such that E D [i2I Ei andEi Ej and LjdEi D Li for all i; j 2 I, j � i.

Clearly, for such a net .Li/i2I there is a unique well-defined linear functional Lon E such that L. f / D Li. f / if f 2 Ei; i 2 I; we shall write L D limi Li.

Lemma 1.18 Let E be an adapted subspace of C.X IR/ and let .�i/i2I be a netof measures �i 2 MC.X / which converges vaguely to � 2 MC.X /. Suppose that


for each i 2 I there is a linear subspace Ei L1.X ; �i/ of E such that .L�i/i2Iis a directed net of linear functionals on E, where L�i . f / WD R

f d�i for f 2 Ei.Then limi L�i D L�.

Proof Let L WD limi L�i . Take h 2 EC. Since E D [i2IEi, h 2 Ei0 for some i0 2 I.Let 2 U , where U is defined by (1.2). Then, by the definition of L,

Zh d�j �

Zhd�j D L�j .h/ D L.h/ for j 2 I; j � i0:

Since h 2 Cc.X ;R/, limiRh d�i D

Rh d� by the vague convergence, so that

Zh d�D sup

2U

Zhd� D sup

2Ulimi

Zhd�i �

Zh d�j D L.h/; j � i0: (1.7)

Thus, h 2 L1.X I�/. Therefore, since E D EC � EC, it follows that E L1.X I�/:Let f 2 EC. There exists a g 2 EC which dominates f , that is, for any " > 0 there

is an h" 2 Cc.X IR/ such that f � "gC h", so that f � h" � "g. Since the index set Iis directed, there is an i0 2 I such that f and g are in Ei0 . Suppose that i 2 I, i � i0.Then, L. f / D R

f d�i by the definition of L and similarly L.g/ D Rg d�i. Using

these facts and (1.7), applied first with h D f and then twice with h D g, we deriveˇˇL. f /�

Zf d�

ˇˇ D L. f /�

Zf d� D

Zf d�i �

Zf d�

DZ. f � h"/ d�i �

Z. f � h"/ d�C

Zh" d�i �

Zh" d�

� "�Z

g d�i CZ

gd�

�CZ

h" d�i �Z

h" d�

� ".L.g/C L.g//CZ

h" d�i �Z

h" d�; i � i0:

Passing to the limit limi the preceding inequality yields jL. f / � R f d�j � 2 "L.g/.Now, letting " ! C0, we get L. f / D R

f d�. Since E D EC � EC, it follows thatL.h/ D R h d� for all h 2 E. utTheorem 1.19 Suppose that L is a moment functional on an adapted subspace E ofC.X IR/. The set ML of representing measures is convex and vaguely compact.

Proof It is clear that ML is convex. Let f 2 Cc.X ;R/. By Lemma 1.6, there existsa g 2 E such j f .x/j � g.x/ for x 2 X . Then

sup�2ML

ˇˇ Z f d�

ˇˇ �

Zg d� D L.g/ <1:

Hence, by Theorem A.6, ML is relatively vaguely compact in MC.X /: It thereforesuffices to show that ML is closed in MC.X / with respect to the vague topology.


For let .�i/i2I be a net from ML which converges vaguely to � 2 MC.X /. FromLemma 1.18, applied with Ei D E for all i, it follows that L� D limi L�i . Since�i 2ML, we have L�i D L. Hence L� D L, that is, � 2ML. utTheorem 1.20 Suppose that X is a locally compact Hausdorff space such thatC0.X IR/, equipped with the supremum norm, is separable. Let E be an adaptedsubspace of C.X IR/ and let .�n/n2N be a sequence of measures �n 2 MC.X /:Suppose for n 2 N there is a linear subspace En L1.X ; �i/ of E such that.L�n/n2N is a directed sequence of functionals on E according to Definition 1.17,where L�n. f / D R f d�n; f 2 En.

Then there exists a subsequence .�nk /k2N that converges vaguely to a measure� 2 MC.X / and we have limk!1 L�nk D L�. If the functional L� is determinate,then the sequence .�n/n2N itself converges vaguely to �.

Proof We first show that the set M WD f�n W n 2 Ng is relatively compact in thevague topology. Let f 2 Cc.X ;R/. By Lemma 1.6, j f .x/j � g.x/ for some g 2 EC.By Definition 1.17, there is a k 2 N such that g 2 Ek and

Rg d�n D

Rg d�k for

n � k. Hence

supn2N

ˇˇZ

f d�n

ˇˇ � sup

n2N

Zg d�n D max

jD1;:::;k

Zg d�j <1;

so M is relatively vaguely compact by Theorem A.6. Further, since C0.X IR/ isseparable, so is its subset Cc.X IR/ and the vague topology on MC.X / is metrizableby Proposition A.7. Therefore, .�n/n2N has a subsequence .�nk /k2N that convergesvaguely to some measure � 2 MC.X /. Then L� D limk!1 L�nk by Lemma 1.18.

Suppose that L� is determinate. Let .�mk /k2N be another subsequence whichconverges vaguely, say to Q�. From Definition 1.17 it follows that limk!1 L�mk isindependent of the subsequence. Therefore, by Lemma 1.18, L Q� D limk!1 L�nk DL� and hence Q� D �, because L� is determinate. Thus, since each convergentsubsequence has the same limit, the sequence .�n/n2N itself converges vaguely. ut

The next result characterizes the extreme points of the convex set ML.

Theorem 1.21 Let E be a linear subspace of C.X IR/. Suppose that E contains afunction e such that e.x/ � 1; x 2 X . Let L be a moment functional on E. Then ameasure � 2ML is an extreme point ofML if and only if E is dense in L1.X I�/.Proof First we assume that � 2ML is not an extreme point of ML. Then there aremeasures �1; �2 2 ML, �1 ¤ �2, such that � D 1

2.�1 C �2/. Since �1 � 2�,

the Radon–Nikodym theorem (Theorem A.3) implies that d�1 D gd� for someg 2 L1.X I�/. By �1 ¤ �2, we have 1� g ¤ 0. Using that �1; � 2ML we obtain

Zf .1 � g/d� D

Zfd� �

Zfgd� D

Zfd� �

Zfd�1 D 0; f 2 E:


Thus, 1�g ¤ 0 defines a nontrivial continuous linear functional on L1.X I�/ whichannihilates E. Hence E is not dense in L1.X I�/.

Conversely, assume that E is not dense in L1.X ; �/. Since e 2 E L1.X ; �/, themeasure � is finite, so L1.X ; �/ is the dual of L1.X ; �/. Hence there is a functiong 2 L1.X ; �/, kgk1 D 1, which annihilates E. Define �˙ by d�˙ D .1˙ g/d�.Since kgk1 D 1, �C and �� are positive measures. From the relation

Rfgd� D 0

for f 2 E it follows that�C and�� are in VL. Since� D 12.�CC��/ and�C ¤ ��

(by g ¤ 0), � is not an extreme point of ML. utRemark 1.22

1. The proof and the assertion of Proposition 1.21 remain valid for Borel functionsrather than continuous functions.

2. If � 2 ML and E is dense in L2.X I�/ in Theorem 1.21, E is also dense inL1.X ; �/, so � is an extreme point of ML. The converse is not true, that is, thereare extreme points � of ML for which E is not dense in L2.X ; �/. ıThe following simple fact will often be used to localize the support of represent-

ing measures. We will apply it mainly to semi-algebraic sets and polynomials.

Proposition 1.23 Let f 2 C.X IR/ and � 2 MC.X /. Suppose that f .x/ � 0 forx 2 X and

Rf .x/ d� D 0. Then

supp� Z. f / WD fx 2 X W f .x/ D 0g:

Proof Let x0 2 X . Suppose that x0 … Z. f /. Then f .x0/ > 0. Since f is continuous,there are an open neighbourhood U of x0 and an " > 0 such that f .x/ � " on U.Then

0 DZXf .x/ d� �

ZUf .x/ d� � "�.U/ � 0;

so that �.U/ D 0. Therefore, since U is an open set containing x0, it follows fromthe definition of the support that x0 … supp�. ut

1.2 Integral Representations of Functionals onFinite-Dimensional Spaces

In this section we suppose that E is a finite-dimensional linear subspace ofC.X IR/: We denote by lx the point evaluation at x 2 X , that is, lx. f / D f .x/for f 2 E.

For some results we will use the following condition:

There exists a function e 2 E such that e.x/ � 1 for x 2 X : (1.8)

1.2 Integral Representations of Functionals on Finite-Dimensional Spaces 23

1.2.1 Atomic Measures

Since E is finite-dimensional, it is natural to look for integral representations offunctionals by finitely atomic measures. To simplify the formulation of the resultswe consider the zero measure as a 0-atomic measure.

Our first main result is the following Richter–Tchakaloff theorem.

Theorem 1.24 Suppose that .Y; �/ is a measure space, V is a finite-dimensionallinear subspace of L1R.Y; �/, and L� denotes the linear functional on V definedby L�. f / D R

f d�, f 2 V. Then there is a k-atomic measure � D PkjD1mjıxj 2

MC.Y/, where k � dimV, such that L� D L� , that is,

Zfd� D

Zfd� �

kXjD1

mj f .xj/; f 2 V:

Proof Let C be the convex cone in the dual space V� of all nonnegative linearcombinations of point evaluations lx, where x 2 Y; and let C be the closure of C inV�. We prove by induction on m WD dim V that L� 2 C:

First let m D 1 and V D R�f . Set c WD Rfd�. If c D 0, then

R.�f /d� D

0 � lx1 .�f /; � 2 R, for any x1 2 Y . Suppose now that c > 0. Then f .x1/ > 0 forsome x1 2 Y . Hence m1 WD cf .x1/�1 > 0 and

R.�f /d� D m1lx1 .�f / for � 2 R.

The case c < 0 is treated similarly.Assume that the assertion holds for vector spaces of dimension m�1: Let V be

a vector space of dimension m. By standard approximation ofRfd� by integrals of

simple functions it follows that L� 2 C. We now distinguish between two cases.Case 1: L� is an interior point of C.

Since C and C have the same interior points (by Proposition A.33(ii)), we haveL� 2 C.

Case 2: L� is a boundary point of C.Then there exists a supporting hyperplane F0 for the cone C at L� (by Proposi-tion A.34(ii)), that is, F0 is a linear functional on V� such that F0 ¤ 0, F0.L�/ D 0and F0.L/ � 0 for all L 2 C. Because V is finite-dimensional, there is a functionf0 2 V such that F0.L/ D L. f0/; L 2 V�. For x 2 Y , we have lx 2 C and henceF0.lx/ D lx. f0/ D f0.x/ � 0: Clearly, F0 ¤ 0 implies that f0 ¤ 0. We choose an.m�1/-dimensional linear subspace V0 of V such that V D V0 ˚ R�f0. Let us setZ WD fx 2 Y W f0.x/ D 0g. Since 0 D F0.L�/ D L�. f0/ D

Rf0d� and f0.x/ � 0

on Y , it follows that f0.x/ D 0 �-a.e. on Y , that is, �.YnZ/ D 0. Now we define ameasure Q� on Z by Q�.M/ D �.M \Z/. Then

L�.g/ DZYg d� D

ZZg d� D

ZZg d Q� D L Q�.g/ for g 2 V0:


We apply the induction hypothesis to the functional L Q� on V0 L1.Z; Q�/. SinceL� D L Q� on V0, there exist �j � 0 and xj 2 Z; j D 1; : : : ; n, such that for f 2 V0,

L�. f / DnX

jD1�jlxj. f /: (1.9)

Since f0 D 0 on Z , hence f0.xj/ D 0, and L�. f0/ D 0, (1.9) holds for f D f0 as welland so for all f 2 V . Thus, L� 2 C. This completes the induction proof.

The set C is a cone in the m-dimensional real vector space V�. Since L� 2 C,Carathéodory’s theorem (Theorem A.35(ii)) implies that there is a representation(1.9) with n � m. This means that L� is the integral of the measure � DPn

jD1 �jıxj .Clearly, � is k-atomic, where k � n � m. (We only have k � n, since some numbers�j in (1.9) could be zero and the points xj are not necessarily different.) ut

The next corollary will be crucial for the study of truncated moment problems.

Corollary 1.25 Each moment functional on a finite-dimensional linear subspace Eof C.X IR/ has a k-atomic representing measure �, where k � dim E. Further, if �is a representing measure of L and Y is a Borel subset of X such that �.XnY/ D 0,then all atoms of � can be chosen from Y .

Proof Apply Theorem 1.24 to the measure space .Y; �dY/ and V D E. utLet L denote the cone of all moment functionals on E. The first assertion of the

following theorem is the counterpart of Proposition 1.9 for atomic measures.

Theorem 1.26 Suppose that X is compact and condition (1.8) is satisfied.

(i) For each EC-positive linear functional L0 on E there exists a k-atomic measure� 2 MC.X /, k � dimE, such that L0. f / D

Rf d� for f 2 E.

(ii) The cone L is closed in the norm topology of the dual space E� of E.

Proof Set m WD dimE. As in the preceding proof, let C denote the cone in the dualspace E� of all nonnegative linear combinations of point evaluations lx at x 2 X . ByCarathéodory’s theorem (Theorem A.35(i)), each L 2 C is a combination of at mostm point evaluations lx, that is, L is of the form

L DmXjD1

�jlxj ; where �j � 0; xj 2 X : (1.10)

We prove that C is closed in E�. Let .L.n//n2N be a sequence from C convergingto L 2 E�. Let �.n/j and x.n/j be the corresponding numbers resp. points in (1.10).Now we use the function e occurring in condition (1.8) and obtain

L.n/.e/ DX

i�.n/i e.x.n/i / � �.n/j ; j D 1; : : : ;m:


Hence, since the converging sequence .L.n/.e//n2N is bounded, so are the sequences.�.n/j /n2N, j D 1; : : : ;m. Because X is compact, there are subsequences .�.nk/j /k2N

and .x.nk/j /k2N which converge in R and X , respectively. Passing to the limit in therepresentation (1.10) for L.nk/ yields L 2 C. Thus, C is closed in E�.

Next we show that L0 2 C. Assume the contrary. Since C is closed, by theseparation of convex sets (Theorem A.26 (ii)) there exists a linear functional F0 onE� such that F0.L0/ < 0 and F0.L/ � 0 for L 2 C. Because E is finite-dimensional,there is an f0 2 E such that F0.L/ D L. f0/; L 2 E�. For x 2 X , lx 2 C and henceF0.lx/ D lx. f0/ D f0.x/ � 0; so that f0 2 EC. But F0.L0/ D L0. f0/ < 0 whichcontradicts the assumption that L0 is EC-positive. This proves that L0 2 C.

Hence L0 is of the form (1.10) and so the integral of the measure� DPmjD1 �jıxj .

Since some �j may be zero, � is k-atomic with k � m. This proves (i).Each functional of L is obviously EC-positive and hence in C by (i). Thus

L C. Since trivially C L, we have L D C. Hence L is closed in E�: Thisproves (ii). ut

Let L denote the closure of the cone L of moment functionals in E�. The nextproposition reformulates some results in terms of dual cones (see (A.19)).

Proposition 1.27 L .EC/^ D L and L^ D EC D .EC/^^:

If K is compact and condition (1.8) is satisfied, then L D .EC/^.

Proof First we prove that .EC/^ L. Assume to the contrary that there existsa functional L0 2 .EC/^ such that L0 … L. Then, by the separation of convexsets (Theorem A.26(i)) applied to the closed (!) cone L in E�, there is a linearfunctional F on E� such that F.L0/ < 0 and F.L/ � 0 for L 2 L^. Since E isfinite-dimensional, there exists an element f 2 E such that F.L/ D L. f /, L 2 E�.For x 2 X , lx 2 L, so that F.lx/ D lx. f / D f .x/ � 0. Hence f 2 EC. Therefore,since L0 2 .EC/^, we get F.L0/ D L0. f / � 0, a contradiction. Thus we have shownthat .EC/^ L.

Since L .EC/^ L as just proved and .EC/^ is obviously closed, .EC/^ DL.

Because EC is closed in E, it follows from the bipolar theorem (Proposition A.32)that EC D .EC/^^: Hence EC D ..EC/^/^ D .L /^ D L^.

Clearly, if L 2 L and p 2 EC, then L. p/ � 0. Therefore, L .EC/^.Now suppose that K is compact and (1.8) holds. Let L 2 .EC/^. Because L is

EC-positive, we have L 2 L by Proposition 1.9 (or by Theorem 1.26(i)). That is,.EC/^L. Since L .EC/^ as noted in the preceding paragraph, L D .EC/^. ut

1.2.2 Strictly Positive Linear Functionals

In this subsection we derive a number of results on strictly EC-positive functionalsthat do not hold for EC-positive functionals in general.


Definition 1.28 A linear functional L on E is called strictly EC-positive if

L. f / > 0 for all f 2 EC; f ¤ 0: (1.11)

The following simple lemma gives some equivalent conditions.

Lemma 1.29 For a linear functional L on E the following are equivalent:

(i) L is strictly EC-positive.(ii) Let k � k be a norm on E: There exists a number c > 0 such that

L. f / � ck fk for f 2 EC: (1.12)

(iii) L is an inner point of the cone .EC/^ in E�.

Proof (i)!(ii) Let c be the infimum of L. f / on the set UC D f f 2 EC W k fk D 1g.For x 2 X , the linear functional lx is continuous on the finite-dimensional normedspace .E; k � k/, so there exists a number Cx > 0 such that

jlx. f /j D j f .x/j � Cxk fk for f 2 E: (1.13)

This implies that EC is closed. Therefore, UC is bounded and closed, hencecompact, in the normed space .E; k � k/. Since L is also continuous on .E; k � k/,the infimum is attained at f0 2 UC. From f0 2 UC we have f ¤ 0 and f 2 EC, sothat L. f0/ D c > 0 by (i). Hence L. f / � c for f 2 UC. By scaling this yields (1.12).

(ii)!(iii) Let k � k� denote the the dual norm of k � k on E�. Suppose that L0 2 E�and kL � L0k� < c. Let f 2 EC Then, using (1.12) we obtain

ck fk � L0. f / � L. f / � L0. f / � kL � L0k� k fk � ck fk

so that L0. f / � 0. Thus, L1 2 .EC/^: This shows that L is an inner point of .EC/^.(iii)!(i) Let f 2 EC; f ¤ 0. Then there exists an x 2 X such that f .x/ > 0. Since

L is an inner point of .EC/^, there exists an " > 0 such that .L�"lx/ 2 .EC/^. HenceL. f / � "f .x/ > 0. ut

In general, EC-positive functionals are not moment functionals (see Exam-ple 1.32), but strictly EC-positive functionals are, as shown by the next theorem.

Theorem 1.30 Let L be a strictly EC-positive linear functional on E.

(i) L is a moment functional.(ii) For each x 2 X , there is a finitely atomic measure � 2 ML such that

�.fxg/ > 0:(iii) For each x 2 X , the infimum

L.x/ WD inf fL. f / W f 2 EC; f .x/ D 1g (1.14)

is a minimum.


Proof

(i) The strictly EC-positive functional L is an inner point of .EC/^ D L byLemma 1.29 and Proposition 1.27. Since the convex set L and its closure Lhave the same inner points by Proposition A.33, L is also an inner point of L.In particular, L belongs to L; that is, L is a moment functional which proves (i).

Let k � k be a norm on E and x 2 X . By Lemma 1.29 and its proof, thereexist positive numbers c and Cx such that (1.12) and (1.13) hold.

(ii) Choose " > 0 such that "Cx < c. Let f 2 EC; f ¤ 0. By (1.12) and (1.13),

.L � "lx/. f / � ck fk � "f .x/ � .c � "/k fk > 0:

Therefore, L � "lx is also strictly EC-positive and hence a moment functionalby (i). Corollary 1.25 implies that L � "lx has a finitely atomic representingmeasure �. Then � WD .�C "ıx/ 2ML is finitely atomic and �.fxg/ � " > 0:

(iii) By (1.14), there is a sequence . fn/n2N of functions fn 2 EC; fn.x/ D 1;

such that limn L. fn/ D L.x/. Since k fnk � c�1L. fn/ by (1.12), . fn/n2N isa bounded sequence in the finite-dimensional normed space .E; k � k/. Henceit has a convergent subsequence . fnk /: Set f D limk fnk : From (1.13) it followsthat f 2 EC and f .x/ D 1. Clearly, L. f / D limk L. fnk / D L.x/, so the infimumin (1.14) is attained at f . ut

Corollary 1.31 Let L be a moment functional on E. Suppose that there exist aclosed subset U of X and a measure � 2 ML such that supp� U and thefollowing holds: If f .x/ � 0 on U and L. f / D 0 for some f 2 E, then f D 0 on U .

Then each x 2 U is an atom of some finitely atomic representing measure of L.

Proof Being a closed subset of X , U is a locally compact Hausdorff space.Since supp� U , there is a well-defined (!) moment functional QL on the linearsubspace QE WD EdU of C.U IR/ given by QL. f dU/ D L. f /; f 2 E. Clearly, QL is. QE/C-positive on QE. The condition on U implies that QL is even strictly . QE/C-positive.Hence the assertion follows from Theorem 1.30(i), applied to QL and QE C.U ;R/.

utExample 1.32 Set X D R and E D Linf1; x; x3; : : : ; x2nC1g, where n 2 N0: Let Lbe a linear functional on E. Then EC D RC � 1. Therefore, if L.1/ > 0, then L isstrictly EC-positive, so that L 2 L by Theorem 1.30(i). If L.1/ D 0 and L ¤ 0, thenL is EC-positive, but L … L:

In the case E D Linfx; x3; : : : ; x2nC1g we have EC D f0g, so each linearfunctional on E is strictly EC-positive and hence E� D L: ı

1.2.3 Sets of Atoms and Determinate Moment Functionals

In this subsection, L denotes a moment functional on E and we suppose that:

For each x 2 X there exists a function fx 2 EC such that fx.x/ > 0: (1.15)


Clearly, condition (1.8) implies (1.15).

Definition 1.33 W.L/ WD fx 2 X W �.fxg/ > 0 for some � 2ML g:That is, W.L/ is the set of points of X that occur as an atom of some representing

measure of L. This is an important set for the moment functionals L.

Lemma 1.34

(i) If L D 0 and � 2ML, then � D 0.(ii) The set W.L/ is not empty if and only if L ¤ 0.Proof

(i) Since L D 0, we have L. fx/ D 0, where fx 2 EC is the function fromassumption (1.15). Hence, by Proposition 1.23, supp� Z. fx/ for all x 2 X .But \x2XZ. fx/ is empty by (1.15), hence is supp�. Thus, � D 0:

(ii) By Corollary 1.25, L has a finitely atomic representing measure �. If L ¤ 0,then � ¤ 0, so W.L/ is not empty. If L D 0, then � D 0 by (i), so W.L/ isempty. ut

Lemma 1.35

(i) Suppose that M is a Borel subset of X containing W.L/. Then �.XnM/ D 0

for each measure � 2ML.(ii) If W.L/ is finite, there exists a measure � 2ML such that supp� D W.L/.

(iii) If W.L/ is infinite, then for any n 2 N there exists a measure � 2ML suchthat jsupp�j � n.

Proof The proofs of all three assertions make essentially use of Theorem 1.24.

(i) Assume �.XnM/ > 0 to the contrary. We define functionals L1 and L2 by

L1. f / DZMf .x/d� and L2. f / D

ZX nM

f .x/d� ; f 2 E:

We apply Theorem 1.24 to the functional L2 on the measure space XnM withmeasure induced from �. Therefore, L2 has a finitely atomic representingmeasure �2 with atoms in XnM. The measure �1 on M which is induced from� is a representing measure of L1. Since � 2ML, we have L D L1 C L2 andhence Q� WD .�1 C �2/ 2 ML: From �.XnM/ > 0 and Lemma 1.34(i) itfollows that L2 ¤ 0. Hence �2 ¤ 0, so there exists an atom x0 2 XnM of�2. Then, Q�.fx0g/ � �2.fx0g/ > 0, that is, x0 2 W.L/ M. This contradictsx0 2 XnM:

(ii) Let W.L/ D fx1; : : : ; xng. By the definition of W.L/, for each xi there is ameasure �i 2ML such that �i.fxig/ > 0. Then � WD 1

n .�1C � � �C�n/ 2ML

and �.fxig/ � 1n�i.fxig/ > 0, i D 1; : : : ; n. Thus, W.L/ supp�. Since

supp� W.L/ by (i) applied with M D W.L/, we have supp� D W.L/.(iii) is proved by a similar reasoning as (ii). utBy Theorem 1.49 below, W.L/ is a closed subset, hence a Borel subset, of X .


Recall from Definition 1.1 that a moment functional L is called determinate if ithas a unique representing measure, or equivalently, if the set ML is a singleton. Thefollowing theorem characterizes determinacy in terms of the size of the set W.L/.

Let f f1; : : : ; fmg be a vector space basis of E. We abbreviate

s.x/ D . f1.x/; : : : ; fm.x//T 2 Rm: (1.16)

Note that s is the moment vector of the delta measure ıx at x 2 X .

Theorem 1.36 For each moment functional L on E the following are equivalent:

(i) L is not determinate.(ii) The set fs.x/ W x 2 W.L/g is linearly dependent in Rm.

(iii) jW.L/j > dim.EdW.L//:(iv) L has a representing measure � such that jsupp�j > dim.EdW.L//:Proof (i)!(iii) Assume to the contrary that jW.L/j � dim.EdW.L// and let �1and �2 be representing measures of L. Then, since dimE is finite, so is W.L/, sayW.L/ D fx1; : : : ; xng with n 2 N. In particular, W.L/ is a Borel set. Hence, fromLemma 1.35(i) applied to M D W.L/; we deduce that supp�i W.L/ for i D 1; 2,so there are numbers cij � 0 for j D 1; : : : ; n; i D 1; 2, such that

L. f / DZ

f .x/ d�i DnX

jD1f .xj/cij for f 2 E:

From the assumption jW.L/j � dim.EdW.L// it follows that there are functionsfj 2 E such that fj.xk/ D ıjk: Then L. fj/ D cij for i D 1; 2, so that c1j D c2j for allj D 1; : : : ; n. Hence �1 D �2, so L is determinate. This contradicts (i).

(iii)!(ii) Since the cardinality of the set fs.x/ W x 2 W.L/g exceeds thedimension of EdW.L/ by (iii), the set must be linearly dependent.

(ii)!(i) Since the set fs.x/ W x 2 W.L/g is linearly dependent, there are pairwisedistinct points x1; : : : ; xk 2 W.L/ and real numbers c1; : : : ; ck, not all zero, such thatPk

iD1 cis.xi/ D 0: Then, since f f1; : : : ; fmg is a basis of E, we have

kXiD1

ci f .xi/ D 0 for f 2 E: (1.17)

We choose for xi 2 W.L/ a representing measure �i of s such that xi 2 supp�i.Clearly, � WD 1

k

PkiD1 �i is a representing measure of s such that �.fxig/ > 0 for all

i. Let " D min f�.fxig/ W i D 1; : : : ; kg. For each number c 2 .�"; "/,

�c D �C c �kX

iD1ciıxi

is a positive (!) measure which represents L by (1.17). By the choice of xi; ci, thesigned measure

Pi ciıxi is not the zero measure. Therefore, �c ¤ �c0 for c ¤ c0.

This shows that L is not determinate.


(iii)$(iv) If W.L/ is finite, by Lemma 1.35(ii) we can choose a � 2 ML suchthat supp� D W.L/. If W.L/ is infinite, Lemma 1.35(iii) implies that there exists a� 2ML such that jsupp�j > dim.EdW.L//: Thus, in both cases, (iii)$(iv). ut

An immediate consequence of Theorem 1.36 is the following.

Corollary 1.37 Let X , E, and L be as in Theorem 1.36. If jW.L/j > dimE or ifthere is a measure� 2ML such that jsupp�j > dimE, then L is not determinate. Inparticular, L is not determinate if W.L/ is an infinite set or if L has a representingmeasure of infinite support.

Corollary 1.38 Suppose that L is a strictly EC-positive moment functional on E.Then L is determinate if and only if jX j � dim E.

Proof From Theorem 1.30(ii), X D W.L/. Therefore, dimE D dim .EdW.L//.Hence the assertion follows from Theorem 1.36(iii)$(i). ut

1.2.4 Supporting Hyperplanes of the Cone of MomentFunctionals

In this subsection, L is a moment functional on E:Now we introduce two other important sets for the moment functional L.

Definition 1.39

NC.L/ WD f f 2 EC W L. f / D 0 g;VC.L/ WD fx 2 X W f .x/ D 0 for f 2 NC.L/ g:

The next proposition contains some properties of these sets.

Proposition 1.40

(i) For each measure � 2ML we have supp� VC.L/.(ii) W.L/ VC.L/:

(iii) If L is strictly EC-positive, then NC.L/ D f0g and VC.L/ D W.L/ D X .

Proof

(i) Since E C.X IR/, Proposition 1.23 implies that supp� VC.L/.(ii) Let x 2 W.L/: Then, by definition, �.fxg/ > 0 for some � 2 ML. Thus

x 2 supp� and hence x 2 VC.L/ by (i).(iii) By Definition 1.28, NC.L/ D f0g. Hence VC.L/ D X . From Theorem 1.30(ii)

we obtain W.L/ D X . utExample 1.41 (An example for which W.L/ ¤ VC.L/) Let X be the subspace ofR2 consisting of three points .�1; 0/; .0; 0/; .1; 0/ and two lines f.t; 1/W t 2 Rg andf.t;�1/W t 2 Rg and let E D RŒx1; x2�2dX : We easily verify that the restriction map


f 7! f dX on RŒx1; x2�2 is injective; for simplicity we write f instead of f dX forf 2 RŒx1; x2�2:

We consider the moment functional L defined by

L. f / D f .�1; 0/C f .1; 0/; f 2 E: (1.18)

We show that NC.L/ D fx2.bx2 C c/W jcj � b; b; c 2 Rg: It is obvious that thesepolynomials are in NC.L/. Conversely, let f 2 NC.L/. Then f .�1; 0/ D f .1; 0/ D 0,so that f D x2.ax1Cbx2Cc/Cd.1�x21/, with a; b; c; d 2 R. Further, d D f .0; 0/ � 0:From f .t;˙1/ � 0 for all t 2 R we conclude that d D 0 and jcj � b.

The zero set VC.L/ of NC.L/ is the intersection of X with the x1-axis, that is,VC.L/ D f.�1; 0/; .0; 0/; .1; 0/g. Let � be an arbitrary representing measure of L:Then, since � is supporting on VC.L/, there are numbers ˛; ˇ; � � 0 such that� D ˛ı.�1;0/Cˇı.0;0/C�ı.1;0/. By (1.18), we have L.x1/ D 0 D

Rx1d� D �˛C�

and L.x21/ D 2 DRx21d� D ˛C �; which implies that ˛ D � D 1. Therefore, since

L.1/ D 2 D R 1d� D ˛CˇC � , it follows that ˇ D 0. Hence, �.f.0; 0/g/ D 0; sothat .0; 0/ … W.L/. Thus, W.L/ ¤ VC.L/. The functional L on E is determinate. ı

If L D 0, then NC.L/ D E and VC.L/ D ;. If L ¤ 0 is a moment functional,then VC.L/ ¤ ;; since it contains the support of all representing measures.

Proposition 1.42

(i) Let p 2 NC.L/; p ¤ 0. Then 'p.L0/ D L0. p/;L0 2 E�, defines a supportingfunctional 'p of the cone L at L. Each supporting functional of L at L is ofthis form.

(ii) L is a boundary point of the cone L if and only if NC.L/ ¤ f0g:(iii) L is an inner point of the cone L if and only if NC.L/ D f0g:Proof

(i) Let p 2 NC.L/; p ¤ 0. Since p 2 EC, the functional 'p is L-positive. Further,'p.L/ D L. p/ D 0. Since p ¤ 0, there exists an x 2 X such that p.x/ ¤ 0.Then 'p.lx/ D lx. p/ D p.x/ ¤ 0, so that 'p ¤ 0: This shows that 'p is asupporting functional of L at L.

Conversely, let ' be a supporting functional of L at L. Since E is finite-dimensional, we have .E�/� Š E, so ' D 'p for some p 2 E. For x 2 X , lx 2 Land hence '.lx/ D lx. p/ D p.x/ � 0. That is, p 2 EC. From '.L/ D L. p/ D 0we obtain p 2 NC.L/. Clearly, p ¤ 0, because ' ¤ 0.

(ii) By Proposition A.34(ii), L is a boundary point of L if and only if there is asupporting functional of L at L. Hence (i) implies the assertion of (ii).

(iii) follows from (ii), since L is inner if and only if it is not a boundary point. utA nonempty exposed face (see Definition A.36) of a cone C in a finite-

dimensional real vector space is a subcone F D f f 2 C W '. f / D 0g for somefunctional ' 2 C^.

By Proposition 1.27, L^ D EC, that is, each functional ' 2 L^ is of the form'p.L0/ D L0. p/;L0 2 E�; for some p 2 EC. Hence the nonempty exposed faces of


the cone L in E� are exactly the sets

Fp WD fL0 2 L W 'p.L0/ � L0. p/ D 0g; where p 2 EC: (1.19)

Let L 2 L. Since L .EC/^, NC.L/ is an exposed face of the cone EC. If X iscompact and condition (1.8) holds, then .EC/^ D L by Proposition 1.27 and henceeach exposed face of EC is of this form. Thus, in this case the subcones NC.L/ areprecisely the nonempty exposed faces of the cone EC.

Proposition 1.43 Let L 2 L and x 2 K: Suppose that X is compact and condition(1.8) is fulfilled. Then the supremum

cL.x/ WD sup fc 2 R W .L � cLx/ 2 L g (1.20)

is attained and we have cL.x/ � e.x/�1L.e/: Further, L � cL.x/lx is a boundarypoint of L and there exists a p 2 EC; p ¤ 0; such that L. p/ D cL.x/p.x/:

Proof If .L � clx/ 2 L, then .L � clx/.e/ � 0, so that c � e.x/�1L.e/. Therefore,c.x/ � e.x/�1L.e/:

Since K is compact and (1.8) holds, the cone L is closed in E� by Theorem 1.26.There is a sequence .cn/n2N such that .L � cnlx/ 2 L for all n and limn cn D cL.x/.Then .L � cnlx/ ! .L � cL.x/lx/. Since L is closed, .L � cL.x/lx/ 2 L, that is, thesupremum (1.20) is attained.

The definition of cL.x/ implies that L � cL.x/lx is a boundary point of L.Therefore, by Proposition 1.42(ii), there exists a p 2 NC.L � cL.x/lx/; p ¤ 0: ThenL. p/ D cL.x/p.x/: utProposition 1.43 is a tool to reduce problems on moment functionals to boundaryfunctionals. If L is an inner point of L, it is clear that cL.x/ > 0 for all x 2 X .

Proposition 1.44 For each moment functional L there exists a p 2 NC.L/ such that

VC.L/ D Z. p/ WD fx 2 X W p.x/ D 0g:

Proof First let L be an inner point of L. Then,NC.L/ D f0g by Proposition 1.42(iii),hence VC.L/ D X ; so we can set p D 0.

Now let L be a boundary point of L. Then, NC.L/ ¤ f0g: Let p1; : : : ; pk be amaximal linearly independent subset of NC.L/. We prove that p WD p1 C � � � C pkhas the desired properties. Obviously, p 2 NC.L/ and VC.L/ Z. p/ by definition.Suppose that x 2 Z. p/. Let q 2 NC.L/. By the choice of p1; : : : ; pk, the function q isa linear combination q DPi �ipi with �i 2 R. From p.x/ D p1.x/C� � �Cpk.x/ D 0and pj.x/ � 0 (by qj 2 NC.L/) it follows that we have pi.x/ D 0 for all i andtherefore q.x/ D 0. Since q 2 NC.L/ was arbitrary, x 2 VC.L/. ut

For inner points of L we have W.L/ D X (by Lemma 1.29 and Theorem 1.30(ii))and hence W.L/ D VC.L/. In general, W.L/ ¤ VC.L/ as shown by Example 1.41.

The next theorem characterizes those boundary points for which W.L/ D VC.L/:


Theorem 1.45 Let L be a boundary point of L. Then W.L/ D VC.L/ if and only ifL lies in the relative interior of an exposed face of the cone L.

Proof First assume that W.L/ D VC.L/: By Proposition 1.44, there exists a p 2NC.L/ such that Z. p/ D VC.L/. Since 'p 2 L^ and L. p/ D 0, the set Fp definedby (1.19) is an exposed face of L containing L. We will prove that L is an inner pointof Fp.

If x 2 Z. p/, then lx. p/ D p.x/ D 0, so that lx 2 Fp. We choose a maximalnumber of points x1; : : : ; xk 2 Z. p/ such that the point evaluations lx1 ; : : : ; lxk onE are linearly independent. Since Z. p/ D VC.L/ D W.L/; we have xi 2 W.L/.Therefore, for each i there exists a representing measure �i of L such that xi is anatom of �i. Then � WD 1

k

PkiD1 �i is also a representing measure of L and each xi is

an atom of �.Suppose that L0 2 Fp. Let�0 DPj cjıyj be a finitely atomic representing measure

of L0. Since L0. p/ D 0, we have supp�0 2 Z. p/, so that yj 2 Z. p/ for all j. Hence,by the choice of the points xi, L0 D P

j cjlyj is in the span of lx1 ; : : : ; lxk . That is,

there are reals �1; : : : ; �k such that L0 DPkiD1 �ilxi . Since � has positive masses at

all points xi, there exists an " > 0 such that �C c � �0 is a positive (!) measure forc 2 .�"; "/. Its moment functional is .L C cL0/ 2 L: Therefore, .L C cL0/ 2 Fp,since L;L0 2 Fp. This shows that L is an inner point of Fp.

Conversely, suppose that L is an inner point of an exposed face F of L. Then Fis of the form (1.19) for some p 2 NC.L/. Let x 2 VC.L/. Then, since p 2 NC.L/,p.x/ D 0 and hence lx 2 F. Since L in an inner point of F, there is a c > 0 suchthat L0 WD L � c � lx 2 F. If �0 is a representing measure of L0, then � D �0 C c � ıxis a representing measure of L and �.fxg/ � c > 0, so that x 2 W.L/. SinceW.L/ VC.L/ always holds by Proposition 1.40(ii), we get W.L/ D VC.L/. ut

Most results of this section are stated in terms of moment functionals. Sometimesit is convenient to work instead with moment sequences and the moment cone.

Fix a vector space basis f f1; : : : ; fmg of E: Let S denote the cone of momentsequences s D .sj/mjD1, that is, of sequences of the form sj D

Rfj.x/ d� for some

Radon measure � on X . The linear map s 7! Ls is a homeomorphism of S Rm

onto L E�: Using this simple fact notions and results on L can be reformulatedin terms of the cone S and vice versa. We encourage the reader to carry this out. Asa sample, we describe the supporting hyperplanes and exposed faces of the cone S:

The vector s.x/ D . f1.x/; : : : ; fm.x//T 2 Rm from (1.16) is just the momentvector of the delta measure ıx, x 2 X . Let h�; �i be the standard scalar product on Rm.

For v D .v1; : : : ; vm/T 2 Rm we abbreviate

fv WD v1 f1 C � � � C vm fm:

Then E D f fv W v 2 Rmg. Since fv.x/ D hv; s.x/i for x 2 X ; we have

EC D f fv W v 2 Rm; hv; s.x/i � 0 for x 2 X g:


It is easily verified that Ls. fv/ D hv; si for v; s 2 Rm: This implies that for eachlinear functional h on E there is a unique vector u 2 Rm such that

hu. fv/ WD h. fv/ D hv; ui; v 2 Rm: (1.21)

Set NC.s/ WD NC.Ls/. For u 2 Rm we define

hu.t/ WD hu; ti; t 2 Rm; and Hu WD fx 2 Rm W hu; xi D 0g:

Lemma 1.46 Let u 2 Rm and s 2 S. Then fu 2 NC.s/ if and only if hu.s/ D 0 andhu.t/ � 0 for t 2 S.

Proof Let t 2 S. Since t has a finitely atomic representing measure, we can writet DPi cis.xi/; where xi 2 X and ci � 0 for all i. Using (1.21) we compute

Lt. fu/ D hu; ti D hu.t/ DX

icihu; s.xi/i D

Xici fu.xi/:

Using the definition of NC.s/ the assertions follow at once from this equality. utLet s 2 S. By Lemma 1.46, the functional hu, u 2 Rm, is a supporting hyperplane

of S at s if and only if fu 2 NC.s/ and u ¤ 0. Each supporting hyperplane of S at sis of the form. Thus, s is a boundary point of S if and only if NC.s/ ¤ f0g.

Further, Hu \ S is an exposed face of S if and only if fu 2 NC.s/. All nonemptyexposed faces of S are of this form.

1.2.5 The Set of Atoms and the Core Variety

Throughout this section, L is a moment functional on E such that L ¤ 0.We define inductively linear subspaces Nk.L/, k 2 N; of E and subsets Vj.L/,

j 2 N0; of X by V0.L/ D X ,

Nk.L/ WD f p 2 E W L. p/ D 0; p.x/ � 0 for x 2 Vk�1.L/ g; (1.22)

Vj.L/ WD fx 2 X W p.x/ D 0 for p 2 Nj.L/g: (1.23)

Definition 1.47 The core variety V.L/ of the moment functional L on E is

V.L/ WD1\jD0

Vj.L/: (1.24)

Since L ¤ 0, it follows from Proposition 1.48(ii) that Vk.L/ ¤ ; for all k 2 N:Note that N1.L/ D NC.L/ and V1.L/ D VC.L/; where NC.L/ and VC.L/ have beenintroduced in Definition 1.39.


Some properties of these sets are contained in the next proposition. Assertion(1.25) is the crucial step in the proof of Theorem 1.49 below.

Proposition 1.48

(i) Nj�1.L/ Nj.L/ and Vj.L/ Vj�1.L/ for j 2 N:

(ii) If � is representing measure of L, then supp� V.L/:(iii) There exists a k 2 N0 such that

X D V0.L/ ¥ V1.L/ ¥ : : : ¥ Vk.L/ D VkCj.L/ D V.L/; j 2 N: (1.25)

Proof

(i) follows easily by induction; we omit the details.Let j 2 N0. We denote by E. j/ WD EdVj.L/ C.Vj.L/IR/ the vector space

of functions fj WD f dVj.L/, where f 2 E, and by L. j/ the corresponding coneof moment functionals on E. j/. Clearly, E.0/ D E and L D L.0/. Note that ingeneral dim E. j/ is smaller than dim E.

(ii) We prove by induction that supp� Vj.L/ for j 2 N0. For j D 0 this isobvious. Assume that this holds for some j. Then

L. j/. fj/ DZVj.L/

f .x/ d�; f 2 E; (1.26)

defines a moment functional on E. j/. Then supp� VC.L. j// D VjC1.L/by Proposition 1.40(i) which completes the induction proof. Thus supp� \j Vj.L/ D V.L/:

(iii) Fix � 2 ML. Let L. j/ 2 L. j/ be given by (1.26). Then, NjC1.L/ D NC.L. j//and VjC1.L/ D VC.L. j//. From Proposition 1.44, applied to L. j/; we concludethat there exists a pjC1 2 E such that pjC1dVj.L/ 2 NC.L. j// D NjC1.L/ and

VC.L. j// D VjC1.L/ D Z. pjC1dVj.L// D fx 2 Vj.L/ W pjC1.x/ D 0g:(1.27)

Suppose that L is an inner point of L. Then, by Proposition 1.42(iii),NC.L/ D N1.L/ D f0g, so that V1.L/ D X . Hence it follows from thecorresponding definitions that Nj.L/ D f0g and Vj.L/ D X for all j 2 N,so the assertion holds with k D 0.

Now let L be a boundary point of L: Then N1.L/ ¤ f0g and hence V1.L/ ¤X . Assume that r 2 N and

V0.L/ ¥ : : : ¥ Vr.L/: (1.28)

We show that the functions p1; : : : ; pr are linearly independent. Assume thecontrary. Then

PrjD1 �jpj D 0, where �j 2 R, not all zero. Let n be the

largest index such that �n ¤ 0. Then pn.x/ D Pj<n �j�

�1n pj. (The sum is

to set zero if n D 1.) Since Vi.L/ Vj.L/ if j � i and pj vanishes on Vj.L/ by


(1.27), it follows that pn D 0 on Vn�1.L/. Hence Vn.L/ Vn�1.L/ by (1.27), acontradiction to (1.28).

In the preceding two paragraphs we have shown that there is a k 2 N0,k � dimE, such that Vk.L/ D VkC1.L/. Then NkC1.L/ D NkC2.L/ and henceVkC1.L/ D VkC2.L/: By induction, VkCj.L/ D Vk.L/ for j 2 N, so that V.L/ DVk.L/. ut

The following result says that the core variety is just the set of possible atoms. Itimplies that W.L/ is a Borel subset of X .

Theorem 1.49 Suppose that L is a moment functional on E and L ¤ 0. ThenW.L/ D V.L/. In particular, W.L/ is a closed subset of X .

Proof From Proposition 1.48(ii) it follows that W.L/ V.L/.By Proposition 1.25, there exists a k 2 N0 such that (1.25) holds. We

show that the set U WD V.L/ fulfills the assumptions of Corollary 1.31. ByProposition 1.48(ii),we have supp� V.L/. Further, if f 2 E satisfies f .x/ � 0

on U D Vk.L/ and L. f / D 0, then f 2 NkC1.L/ and hence f .x/ D 0 onVkC1.L/ D V.L/ D U : Now Corollary 1.31 applies and gives the converse inclusionU D V.L/ W.L/. Thus, W.L/ D V.L/.

Since V.L/ D Vk.L/ is closed by its definition, so is W.L/. ut

1.2.6 Extremal Values of Integrals with Moment Constraints

In this subsection, we investigate the supremum and infimum of the integralRh d�

of some measurable function h under the constraint that the measure � has given“moments”

Rfj d� D sj; j D 1; : : : ; n:

Let M.E/ denote the set of Radon measures � on X for which all functions ofE are �-integrable. We fix a vector space basis f f1; : : : ; fng of the finite-dimensionalsubspace E of C.X IR/ and define the moment cone

S D�s D .s1; : : : ; sn/ W sj D

Zfj.x/d�; j D 1; : : : ; n; where � 2M.E/

�:

For s 2 S let Ms denote the set of representing measures of s, that is,

Ms D f� 2M.E/ W sj DZ

fj.x/d�; j D 1; : : : ; n g:

Let h be a fixed real-valued Borel function on X such that the integralRh d� is

finite for all � 2Ms. For instance, if the function h is bounded on X and condition(1.8) holds, then each measure� 2Mt; t 2 S; is finite and hence

Rh.x/d� is finite.

For an interior point s of S, we are interested in the upper bound Isup.hI s/ andthe lower bound Iinf.hI s/ of the integral

Rh.x/ d� under the constraints � 2Ms:


Isup.hI s/ D sup˚ Z

h.x/d� W � 2Ms�; (1.29)

Iinf.hI s/ D inf˚ Z

h.x/d� W � 2Ms�: (1.30)

If h is the indicator function �A of a Borel set A, Isup.�AI s/ is the supremum ofmasses�.A/ for measures� 2Ms. This is an important quantity in moment theory.

Let Sext denote the moment cone obtained by adding the function h to thesequence f f1; : : : ; fng. That is,

Sext D�.t; tnC1/ 2 RnC1 W t 2 S; tnC1 D

Zh.x/d� for � 2Mt

�:

If h 2 E, thenRh d� D Ls.h/ for � 2 Ss, so that Isup.hI s/ D Iinf.hI s/ D Ls.h/, so

we are done in this case. Further, we set

E�.h/ D f f 2 E W f .x/ � h.x/ for x 2 X g; (1.31)

E�.h/ D f f 2 E W f .x/ � h.x/ for x 2 X g: (1.32)

The following two results relate the problems (1.29) and (1.30) to two otherproblems (1.33) and (1.34) and give existence criteria for these problems.

Theorem 1.50 Let s be an interior point of S such that Isup.hI s/ and Iinf.hI s/ arefinite. Further, assume that

Rh d� is finite for all � 2Mt and t 2 S. Then

Isup.hI s/ D inf fLs. f / W f 2 E�.h/g; (1.33)

Iinf.hI s/ D sup fLs. f / W f 2 E�.h/g; (1.34)

and there exist functions fC 2 E�.h/ and f� 2 E�.h/ such that the infimum in (1.33)is attained at fC and the supremum in (1.34) is attained at f�.

Further, if the supremum in (1.29) is attained at �C 2 Ms and fC 2 E�.h/satisfies Isup.hI s/ D Ls. fC/, then �C is supported on fx 2 X W h.x/ D fC.x/g:

Likewise, if the infimum in (1.30) is attained at some �� 2Ms and f� 2 E�.h/is such that Iinf.hI s/ D Ls. f�/, then �� is supported on fx 2 X W h.x/ D f�.x/g:Proof As stated before the theorem, the assertion holds if h 2 E. Thus we canassume that h … E.

Since Isup.hI s/ 2 R is the supremum (1.29), .s; Isup.hI s// is in the closure of Sext

and .s; Isup.hI s/C "/ … Sext for all " > 0. Thus .s; Isup.hI s// is a boundary point ofthe convex cone Sext of RnC1, so there exists a supporting hyperplane through thispoint at Sext. That is, there are a 2 Rn and anC1 2 R such that .a; anC1/ ¤ 0 and

a � tC anC1tnC1 � 0 for all .t; tnC1/ 2 Sext; (1.35)

a � sC anC1Isup.hI s/ D 0: (1.36)


For any x 2 X , the delta measure ıx is in M.E/. Letting t be the moment sequenceof ıx in (1.35) we obtain

a1 f1.x/C � � � C an fn.x/C anC1h.x/ � 0 for x 2 X : (1.37)

Next we prove that anC1 < 0. Since h … E, it is easily seen that Sext

is not contained in a hyperplane. Hence there is a .t; tnC1/ 2 Sext such thata � tC anC1tnC1 > 0. Since s is an inner point of S, we have t0 WD sC ".s� t/ 2 Sfor small " > 0. Put t0nC1 WD

Rhd� for some � 2 Mt0 . Then .t0; t0nC1/ 2 Sext:

Therefore, setting t00nC1 WD .1C "/�1t0nC1 C ".1C "/�1tnC1,

.s; t00nC1/ D .1C "/�1.t0; t0nC1/C ".1C "/�1.t; tnC1/

is a convex combination of points from Sext, so that .s; t00nC1/ 2 Sext: The inequalitiesa � t C anC1tnC1 > 0 and a � t0 C anC1t0nC1 � 0 imply that a � s C anC1t00nC1 > 0:

Combining the latter with (1.36) we obtain anC1t00nC1 > anC1Isup.hI s/. Therefore,since .s; t00nC1/ 2 Sext and hence Isup.hI s/ � t00nC1 by the definition of Isup.hI s/, itfollows that anC1 < 0.

Set fC.x/ WD �a1a�1nC1f1 � � � � � ana�1

nC1fn. Dividing (1.37) by �anC1 > 0 we getfC.x/ � h.x/ � 0 for x 2 X , so that fC 2 E�.h/. From (1.36) and (1.37) we deriveLs. fC/ D Isup.hI s/: Thus we have shown that the infimum in (1.33) is attained at fCand equal to Isup.hI s/.

The assertion concerning Iinf.hI s/ follows either by a similar reasoning ordirectly from the result on Isup.hI s/, applied with h replaced by �h and fi by �fi.

Finally, let us suppose that the supremum (1.29) is attained for some �C 2Ms

and let Isup.hI s/ D Ls. fC/, where fC 2 E�.h/: Then

ZfC.x/d�C D Ls. fC/ D Isup.hI s/ D

Zh.x/ d�C;

so thatR. fC.x/�h.x// d�C D 0. Since fC�h � 0 on X , it follows that fC�h D 0

�C-a.e.. This means that �C is supported on the set fx 2 X W h.x/ D fC.x/g. Theproof for �� is similar. utRemark 1.51 Theorem 1.50 and its proof remain valid if E consists of measurablefunctions for some �-algebra instead of continuous functions. ı

Since h.x/ � f .x/ on X for f 2 E�.h/, we have

inff2E�.h/

Zjh � f jd� D inf

f2E�.h/

�Zhd��

Zfd�

�DZ

hd�� supf2E�.h/

Ls. f /:

Thus, finding the supremum in (1.34) is equivalent to the problem of finding thebest approximation of h in L1.X ; �/ by functions from E�.h/: In other words,


the supremum in (1.34) is attained at f 2 E�.h/ if and only if f is the bestapproximation of h in L1.X ; �/ by functions of E from below.

Recall that a real-valued function h on X is called upper semicontinuous if foreach a 2 R the set fx 2 X W f .x/ < ag is open in X . Obviously, the characteristicfunction h D �A of some subset A is upper semicontinous if and only if A is closed.

The next theorem is the main result of this section. It contains sufficientconditions ensuring that the supremum in (1.29) is attained.

Theorem 1.52 Suppose that X is a compact topological space and E is a finite-dimensional subspace of C.X IR/ satisfying condition (1.8). Let s be an interiorpoint of S and h an upper semicontinuous bounded real-valued function on X .

Then there exist a k-atomic measure �C D PkjD1mjıxj 2Ms, k � dimE C 1,

and a function fC 2 E�.h/ such that the supremum (1.29) and the infimum (1.33)are attained at fC and �C, respectively, and both numbers coincide. That is, wehave fC.x/ � h.x/ on X , h.xj/ D fC.xj/ for j D 1; : : : ; k, and for each � 2Ms,

Zf .x/d� D

Zf .x/d�C D

kXjD1

mjf .xj/ for f 2 E;

sup�2Ms

Zh.x/d� D

Zh.x/d�C D

kXjD1

mjh.xj/ DZ

fC.x/d� D inff2E�.h/

Zfd�:

Proof From the definition of the supremum (1.29) it follows that there is a sequence.�n/n2N of measures �n 2 Ms such that limn

Rhd�n D Isup.hI s/. By condition

(1.8), we have�n.X / �Red�n D Ls.e/ for n 2 N. Therefore, Theorem A.6 applies

and implies that the set f�n W n 2 Ng is relatively vaguely compact. Then there exista Radon measure �C 2 MC.X / and a subnet .�ni/i2I which converges vaguely to�C. Since X is compact, E Cc.X ;R/ and hence

Rfd�C D limi

Rfd�ni D Ls. f /

for f 2 E. Thus, �C 2Ms.Further, the function 1 is in Cc.X ;R/ again by the compactness of X . Hence

lim �ni .X / D limR1 d�ni D

R1 d� D �.X /, so condition (i) in Proposition A.8

is fulfilled. From�n.X / � Ls.e/ and�.X / � Ls.e/we get�n; � 2 MbC.X /. Since his upper semicontinuous, it follows from the implication (i)!(iii) of Proposition A.8that

Rhd�C � limsupi

Rhd�ni D Isup.hI s/. Obviously, Isup.hI s/ �

Rhd�C by

definition. Thus,Rhd�C D Isup.hI s/, so the supremum (1.29) is attained at �C.

Applying Theorem 1.24 to the functional L. f / D Rf d� on E ˚ R � h

we conclude that �C can be chosen k-atomic, say �C D PkjD1mjıxj , with

k � dimEC 1.All remaining assertions are contained in Theorem 1.50. Because �C is sup-

ported on the set fx 2 X W h.x/ D fC.x/g, we have h.xj/ D fC.xj/ for allj D 1; : : : ; k. ut

In particular, Theorem 1.52 holds if h is the characteristic function of a closedsubset of X . If we assume that the function h is lower semicontinuous (for instance,


the characteristic function of an open set), then the counterpart of Theorem 1.52remains valid almost verbatim for the infimum in (1.30) and the supremum in (1.34).

1.3 Exercises

1. Decide whether or not the following subspace E of C.RIR/ is an adaptedspace:

a. E is the span of functions .x2 C n/�1, n 2 N.b. E is the span of functions e˛x, ˛ > 0.c. E D Cc.RIR/.

2. Are the polynomials RŒx�n of degree at most n an adapted space of C.RIR/ orof C.Œ�1; 1�IR/?

3. Let E be the vector space of bounded continuous real functions on R. Eachf 2 E has a unique continuous extension Of to the Stone–Cech compactificationˇ.R/ of R (see e.g. [Cw, Chapter V, §6]). Let x0 be a point of ˇ.R/nR anddefine L. f / WD Of .x0/, f 2 E. Show that L is an EC-positive linear functional onE which is not a moment functional.

4. Let �n, n 2 N; and � be positive Radon measures on a locally compactHausdorff space X such that the sequence .�n/n2N converges vaguely to �and

P1nD1 �n.X / <1: Show that limn

Rfd�n D

Rfd� for f 2 C0.X IR/.

5. Let a; b 2 R, where a < b. Determine the character set OA for the �-algebrasA D C.RIR/CR � �a and A D C.RIR/CR � �Œa;b�:Hint: Look for a topological space X such that A is isomorphic to C.X ;R/.

In the following exercises, E is a finite-dimensional subspace of C.X IR/, S is themoment cone, and Sj is the set of s 2 S which have a k-atomic representing measurewith k � j.

6. Give an example such that S1 is not closed.7. Prove that Rm D S � S, where m D dim E:8. Let C be the smallest number such that each s 2 S has a k-representing measure

with k � C: Show that Sj�1 ¤ Sj for j D 1; : : : ;C:9. Assume that condition (1.15) holds.

(i) Prove that the cone S is pointed, that is, S \ .�S/ D f0g:(ii) Prove that if S1 is closed, so is Sk for all k 2 N.

10. Assume that X is compact and (1.8) holds. Prove Lk is closed for all k 2 N.11. Suppose that L is a strictly EC-positive moment functional on E and jX j >

dim E. Prove that L is not determinate by using the Hahn–Banach theorem.Hints: Take h 2 Cc.X IR/, h … E: Show that Sh < Ih, where

Sh WD sup fL. f /I f 2 E and f .x/ � h.x/; x 2 X g; (1.38)

Ih WD inf fL. f / W f 2 E and h.x/ � f .x/; x 2 X g: (1.39)

1.4 Notes 41

Choose � 2 R, Sh < � < Ih: Show that the functional L� on F WD E C R � hdefined by L� . f C �h/ D L. f / C ��; f 2 E; is strictly FC-positive. ApplyTheorem 1.30.

1.4 Notes

Choquet’s theory of adapted spaces was elaborated in [Chq]. Haviland’s The-orem 1.12 goes back to [Ha]. Since the one-dimensional case was noted byM. Riesz [Rz2], Theorem 1.12 is often called Riesz–Haviland theorem in theliterature. Theorem 1.21 (for polynomials) is due to M.A. Naimark [Na]; the generalversion is from [Do].

The important Theorem 1.24 was first proved in full generality by H. Richter [Ri]in 1957, see also W.W. Rogosinsky [Rg]. V. Tchakaloff [Tch] treated the simplercompact case about the same time. Richter’s paper has been ignored in the literatureand a number of versions of his result have been reproved even recently.

Theorem 1.26 is due to [FN2]. Assertion (i) of Theorems 1.30 is based on anidea from [FN1]. Theorems 1.30, 1.36, 1.45, and 1.49 were proved in [DSm1]. Thecore variety (for polynomials) was introduced in [F2]. More results on the momentcone can be found in [DSm1], [DSm2]; proofs of Exercises 1.7–1.10 are given in[DSm2].

The results of Sect. 1.2.6 were obtained in [Ri], [Rg], [Ii], [Kp1]; see [Kp2] for asurvey. They will not be used later in this book.

Chapter 2Moment Problems on Abelian �-Semigroups

In this chapter we collect a number of general concepts and simple facts on momentproblems on commutative �-semigroups that will be used throughout the text, oftenwithout mention. Section 2.1 is about positive functionals on �-algebras and positivesemidefinite functions on �-semigroups. In Sect. 2.2 we specialize to commutative�-algebras and �-semigroups and introduce moment functionals, moment functions,K-determinate moment functions, and generalized Hankel matrices. Some standardexamples of commutative �-semigroups are given in Sect. 2.3.

Throughout this chapter, K is either the real field R or the complex field C.

2.1 �-Algebras and �-Semigroups

In this section we discuss the one-to-one correspondence between positive semidef-inite functions on �-semigroups and positive functionals on semigroup �-algebras.

Let us begin with some basic definitions.

Definition 2.1 A �-algebra A over K is an algebra over K equipped with amapping � W A! A, called involution, such that for a; b 2 A and ˛; ˇ 2 K,

.˛aC ˇb/� D ˛ a� C ˇ b�; .ab/� D b�a�; .a�/� D a:

The Hermitian part Ah of a �-algebra A is Ah WD fa 2 A W a D a�g:Our standard examples of real or complex �-algebras in this book are the

polynomial algebras RŒx1; : : : ; xd� and CŒx1; : : : ; xd�, respectively, with involutionsdetermined by .xj/� D xj; j D 1; : : : ; d.


43

44 2 Moment Problems on Abelian �-Semigroups

Definition 2.2 Let A be a K-linear subspace of a some �-algebra over K. We define

A2 D Lin fb�a W a; b 2 Ag; (2.1)

XA2 D

� kXiD1.ai/

�ai W ai 2 A; k 2 N

�: (2.2)

A linear functional L W A2 ! K is called positive if L is .P

A2/-positive, that is, if

L.a�a/ � 0 for a 2 A: (2.3)

Lemma 2.3 Let A be a linear subspace of a �-algebra B over K and L W A2 ! K

a positive linear functional. Then the Cauchy–Schwarz inequality holds:

jL.b�a/j2 � L.a�a/L.b�b/ for a; b 2 A: (2.4)

Further, if B is unital and the unit element of B is contained in A, then

L.a�/ D L.a/ for a 2 A: (2.5)

Proof We carry out the proof in the case K D C; the case K D R is even simpler.For ˛; ˇ 2 K and a; b 2 A we have

L..˛ C ˇb/�.˛aC ˇb//D ˛˛L.a�a/C ˛ˇL.a�b/C ˛ˇL.b�a/C ˇˇL.b�b/ � 0: (2.6)

Hence ˛ˇL.a�b/ C ˛ˇL.b�a/ is real. Letting ˛ˇ D 1 and ˛ˇ D i, we deriveL.a�b/ D L.b�a/. If B has a unit element 1 and 1 2 A, we set b D 1 and get (2.5).

The expression in (2.6) is a positive semidefinite quadratic form. Hence itsdiscriminant has to be nonnegative. Since L.a�b/ D L.b�a/ as just shown, thisyields

L.a�a/L.b�b/� L.b�a/L.a�b/ D L.a�a/L.b�b/� jL.b�a/j2 � 0: ut

If the linear subspace A in Definition 2.2 is itself a �-algebra, then (2.3) is justthe definition of a positive functional on the �-algebra A.

In this book we deal mainly with commutative real algebras. Each such algebrais a �-algebra over R if we take the identity map as involution. In this case,

PA2

is the set of finite sums of squares a2 of elements a 2 A and the Cauchy–Schwarzinequality (2.4) has the following form:

L.ab/2 � L.a2/L.b2/ for a; b 2 A: (2.7)

2.1 �-Algebras and �-Semigroups 45

By a semigroup .S; ı/we mean a nonempty set S with an associative compositionı (that is, a mapping S�S 3 .s1; s2/ 7! s1ıs2 2 S such that s1ı.s2ıs3/ D .s1ıs2/ıs3for all s1; s2; s3 2 S) and a neutral element e 2 S (that is, e ı s D s ı e D s for s 2 S).

Definition 2.4 A �-semigroup .S; ı;�/ is a semigroup .S; ı/ endowed with amapping � W S! S, called an involution, such that

.s ı t/� D t� ı s� and .s�/� D s; s; t 2 S:

If no confusion can arise we write simply S instead of .S; ı;�/.Any abelian semigroup becomes a �-semigroup if we take the identity map as

involution. Each group S is a �-semigroup with involution s� WD s�1, s 2 S.

Definition 2.5 A function ' W S ! K on a �-semigroup S is positive semidefiniteif for arbitrary elements s1; : : : ; sn 2 S, numbers �1; : : : ; �n 2 K and n 2 N,

nXi;jD0

'.s�i ı sj/ �i �j � 0:

The set of positive semidefinite functions ' W S! K on S is denoted by PK.S/.

Suppose that S is �-semigroup. We define the semigroup �-algebra KŒS�. Avector space basis of KŒS� is given by the elements of S and product and involutionof KŒS� are determined by the corresponding operations of S. That is, KŒS� is thevector space of all sums

Ps2S ˛ss, where ˛s 2 K and only finitely many numbers

˛s are nonzero, with pointwise addition and scalar multiplication. The vector spaceKŒS� becomes a unital �-algebra over K with product and involution defined by

�Xs2S ˛ss

��Xt2S ˇtt

� WDPs;t2S ˛sˇt.s ı t/;�X

s2S ˛ss�� WDPs2S ˛s s�:

Since the elements of S form a basis of KŒS�, there is a one-to-one correspon-dence between functions ' W S!K and linear functionals L' W KŒS�!K given by

L'.s/ WD '.s/; s 2 S:

Definition 2.6 The unital �-algebra KŒS� over K is the semigroup �-algebra of Sand the functional L' is called the Riesz functional associated with the function '.

Proposition 2.7 For a function ' W S! K the following are equivalent:

(i) ' is a positive semidefinite function.(ii) L' is a positive linear functional on the �-algebraKŒS�.

(iii) H.'/ D .ast WD '.s� ı t//s;t2S is a positive semidefinite Hermitian matrix.


Proof For arbitrary a DPs2S ˛ss 2 KŒS�, we have

L'.a�a/ D

Xs;t2S '.s

� ı t/˛s˛t DX

s;t2S ast ˛s˛t:

Comparing Definitions 2.5 and 2.2 the first equality implies the equivalence of (i)and (ii), while the second equality yields the equivalence of (i) and (iii). utCorollary 2.8 If ' W S! K is a positive semidefinite function on S, then

'.s�/ D '.s/ and '.s� ı s/ � 0 for s 2 S; (2.8)

j'.s� ı t/j2 � '.s� ı s/'.t� ı t/ for s; t 2 S: (2.9)

In particular, if '.e/ D 0, then '.t/ D 0 for all t 2 S.

Proof By Proposition 2.7, L' is a positive linear functional on the unital �-algebraKŒS�, that is, L'.s�s/ D '.s� ı s/ � 0. Therefore, (2.8) and (2.9) follow at oncefrom (2.5) and (2.4), respectively. If '.e/ D 0, then it follows that ' � 0 on S bysetting s D e in (2.9). utDefinition 2.9 The positive semidefinite matrix

H.'/ D .'.s� ı t//s;t2Swith .s; t/-entry ast WD '.s� ı t/ is called the generalized Hankel matrix associatedwith the positive semidefinite function ' W S! K.

2.2 Commutative �-Algebras and Abelian �-Semigroups

Throughout this section, we assume that A is a commutative unital �-algebra overK and that S is an abelian �-semigroup. As is common, we write C for thecomposition ı of S and denote the neutral element of S by 0.

Definition 2.10 A character on A is linear functional � W A! K satisfying

�.1/ D 1; �.ab/ D �.a/�.b/; �.a�/ D �.a/; a; b 2 A: (2.10)

If A is a real algebra with identity involution, this coincides with Definition 1.13.

The set of characters of A is denoted by OA. We equip OA with the topology ofpointwise convergence and assume that OA is then a locally compact topologicalHausdorff space. The latter is always fulfilled if the algebra A is finitely generated.

The following definition restates Definitions 1.1 and 1.2 in the present setting.

2.2 Commutative �-Algebras and Abelian �-Semigroups 47

Definition 2.11 Let K be a closed subset of OA. A linear functional L W A ! K iscalled a K-moment functional if there exists a Radon measure � on OA such that

supp� K; (2.11)

the function � 7! �.a/ is �-integrable on OA and satisfies

L.a/ DZ

OA�.a/ d�.�/ for all a 2 A: (2.12)

A K-moment functional L is said to be K-determinate if there is only one Radonmeasure � on OA for which (2.11) and (2.12) holds.

In the case K D OA we call K-moment functionals simply moment functionals andK-determinate K-moment functionals determinate.

Lemma 2.12 Each K-moment functional is a positive linear functional on A.

Proof Let a 2 A. For � 2 OA we have �.a�a/ D �.a�/�.a/ D j�.a/j2 by (2.10) andtherefore

L.a�a/ DZ

OA�.a�a/ d�.�/ D

ZOAj�.a/j2 d�.�/ � 0: ut

Remark 2.13 A positive linear functional L on A satisfying L.1/ D 1 is called astate. Let S.A/ denote the set of states of A. Each character of A is an extremepoint of the convex set S.A/. (The reasoning of the proof of Lemma 18.3(ii) belowgives a proof of this well-known fact.) In general not all extreme point of S.A/ arecharacters. (Indeed, by Proposition 13.5, there exists a state L on RŒx1; x2� whichis not a moment functional. From the decomposition theory of states on �-algebras[Sm4, Section 12.4] it follows that L is an integral of extreme points of S.A/. SinceL is not a moment functional, not all extreme points of S.A/ can be characters; seealso the discussion in [Sm4, Remark 12.4.6].) ı

Next we turn to characters on the abelian �-semigroup S.

Definition 2.14 A character of S is a function � W S! K satisfying

�.0/ D 1; �.sC t/ D �.s/�.t/; �.s�/ D �.s/; s; t 2 S: (2.13)

The set S� of characters of an abelian semigroup S is also an abelian �-semigroup, called the dual semigroup of S, with pointwise multiplication ascomposition, complex conjugation as involution and the constant character asneutral element.

Let us assume that S� equipped with the topology of pointwise convergence is alocally compact Hausdorff space. This holds if the �-semigroup is finitely generated.The following is the counterpart of Definition 2.11 for �-semigroups.


Definition 2.15 Let K be a closed subset of S�. We say that a function ' W S! K

is a K-moment function if there exists a Radon measure � on S� such that

supp� K; (2.14)

the function � 7! �.s/ is �-integrable on S� and

'.s/ DZS�

�.s/d�.�/ for all s 2 S: (2.15)

A K-moment function ' is called K-determinate if the measure � satisfying (2.14)and (2.15) is uniquely determined.

If K D S� we call a K-moment function simply a moment function and a K-determinate moment function determinate.

Comparing the preceding definitions and facts we see that we have a one-to-onecorrespondence between notions on a �-semigroup S and its semigroup �-algebraKŒS�: By (2.13) and (2.10), a function � W S! K is a character on the �-semigroupS if and only if its Riesz functional on KŒS� is a character on the �-algebra KŒS�.Comparing Definitions 2.11 and 2.15, it follows that ' is a K-moment functionon S if and only if the Riesz functional L' is a K-moment functional on KŒS�.Further, by these definitions, ' is K-determinate if and only if L' is. That is, themoment problems for the semigroup �-algebra KŒS� and for the �-semigroup S areequivalent. We shall use these fact throughout the book without mention.

From Proposition 2.7 (i)$(ii), and Lemma 2.12 we obtain the following.

Corollary 2.16 Each moment function ' W S! K is positive semidefinite.

By Corollary 2.16 and Lemma 2.12, each moment function ' W S! K is positivesemidefinite and the Riesz functional L' on KŒS� is positive. We shall show later(Proposition 13.5) that the converse is not true for �-semigroup S D Nd

0 when d � 2.Even more, the converse is only true in rare cases! Finding sufficient conditions ona positive linear functional L on RŒNd

0� ensuring that L is a moment functional willbe one of our main tasks in this book.

Next let us suppose that A is a commutative real algebra. We want to defineits complexification AC. The direct sum AC WD A ˚ iA of vector spaces A andiA becomes a commutative complex �-algebra with multiplication, involution andscalar multiplication defined by

.aC ib/.cC id/ D ac� bdCi.bcC ad/; .aC ib/� WD a � ib;

.˛ C iˇ/.aC ib/ WD ˛a � ˇbC i.˛bC ˇa/;

where a; b; c; d 2 A and ˛; ˇ 2 R. This complex �-algebra AC is called thecomplexification of A. The real algebra A is then the Hermitian part .AC/h of AC.

Recall thatP.AC/

2 denotes all finite sumsP

j x�j xj and

PA2 is the set of finite

sumsP

i a2i , where xj 2 AC and ai 2 A.

2.3 Examples 49

Lemma 2.17P.AC/

2 DP A2.

Proof Let x 2 AC. Then x D aCib with a; b 2 A. Using that the algebra A iscommutative (!) we obtain

x�x D .aC ib/�.aC ib/ D a2 C b2 C i.ab� ba/ D a2 C b2 2X

A2: ut

Each R-linear functional L on A has a unique extension LC to a C-linear functionalon AC. An immediate consequence of Lemma 2.17 is the following corollary.

Corollary 2.18 L is positive on the real �-algebra A (with the identity map asinvolution) if and only if LC is positive on the complex �-algebra AC.

At the end of this section we briefly discuss the choice of the field K. A large partof this book deals with the K-moment problem for the �-semigroup S D Nd

0 withidentity involution. By (2.13), all characters on Nd

0 are real-valued, .Nd0/

� Š Rd andwe have RŒNd

0� Š RŒx1; : : : ; xd�, see Example 2.3.1 below. That is, we can workwith the real field and real algebras. In Chaps. 12 and 13 we will apply powerfulmethods from real algebraic geometry to the real algebra A D RŒx1; : : : ; xd�.

But operator-theoretic treatments require complex Hilbert spaces. In this case itis more convenient to use the complex semigroup �-algebra CŒNd

0� Š CŒx1; : : : ; xd�.Since CŒx1; : : : ; xd� is just the complexification of RŒx1; : : : ; xd�, it is easy from thepreceding discussion and Corollary 2.18 to pass from one algebra to the other.

In Chaps. 11 and 15 we will study the moment problem on the unit circle andthe complex moment problem, respectively. In these cases it is unavoidable to workwith the complex field, since otherwise we would not have enough characters.

2.3 Examples

In this section we discuss four important examples that will be crucial for this book.

2.3.1 Example 1: Nd0;n� D n

The additive semigroup Nd0 with identity involution is a �-semigroup and the map

.n1; : : : ; nd/ 7! xn11 � � � xnddis a �-isomorphism of the semigroup �-algebra KŒNd

0� on the �-algebraKŒx1; : : : ; xd� of polynomials with involution determined by x�

j D xj, j D 1; : : : ; d.By identifying .n1; : : : ; nd/ and xn11 � � � xndd we obtain

KŒNd0� Š KŒx1; : : : ; xd�:


Clearly, for any t 2 Rd there is a character �t given by the point evaluation�t. p/ D p.t/; p 2 KŒx1; : : : ; xd�. Conversely, if � is a character, we set tj WD �.xj/for j D 1; : : : ; d. Then �.xj/ D �..xj/�/ D �.xj/, so t D .t1; : : : ; td/ 2 Rd and

�. p.x1; : : : ; xd// D p.�.x1/; : : : ; �.xd// D p.t1; : : : ; td/ D �t. p/:

Hence � D �t. Thus we have shown that the character space of S D Nd0 is

.Nd0/

� D f�t W t 2 Rdg Š Rd; where �t. p/ D p.t/: (2.16)

A function on the semigroup Nd0 is just a multisequence s D .s˛/˛2Nd

0. Its Riesz

functional Ls is given by Ls.x˛/ D s˛ , ˛ 2 Nd0: By the definition of the involution,

positive semidefinite functions on Nd0 are real-valued. By Definition 2.5, a real

sequence s D .s˛/˛2Nd0

is a positive semidefinite function on Nd0 if and only if

X˛;ˇ2Nd

0

s˛Cˇ �˛ �ˇ � 0

for all finite real multisequences .�˛/˛2Nd0:

From Definition 2.15 and Eq. (2.16) it follows that s is a moment function onNd0, briefly a moment sequence, if and only if there is a Radon measure � on Rd Š

.Nd0/

� such that x˛ 2 L1.Rd; �/ and

s˛ DZRd

x˛ d� for ˛ 2 Nd0; (2.17)

or equivalently, p.x/ 2 L1.Rd; �/ and Ls. p/ DRRd p.x/ d� for p 2 RŒx1; : : : ; xd�:

Equation (2.17) means that s˛ is the ˛-th moment of the measure �. Thus, s is amoment sequence if and only if there is a Radon measure � such that each s˛ isthe ˛-th moment s˛.� of �, or equivalently, Ls is a moment functional accordingto Definition 2.11. This explains and justifies the names “moment sequence” and“moment functional”.

By Definition 2.9, the corresponding generalized Hankel matrix H.s/ has the.˛; ˇ/-entry s˛Cˇ . In the case d D 1 the matrix H.s/ is a “usual” one-sided infiniteHankel matrix which is constant on cross-diagonals:

H.s/ D

0BBBBBBB@

s0 s1 s2 : : : sn : : :

s1 s2 s3 : : : snC1 : : :s2 s3 s4 : : : snC2 : : :: : : : : : : : : : : : : : : : : :

sn snC1 snC2 : : : s2n : : :

: : : : : : : : : : : : : : : : : :

1CCCCCCCA: (2.18)

2.3 Examples 51

Remark 2.19

1. It should be emphasized that notions such as positive semidefinite sequence,moment sequence, and Hankel matrix depend essentially on the underlying �-semigroup.

2. In the literature the Hankel matrix is often called the moment matrix, because itsentries are moments if s is a moment sequence. We will not use this terminology.The reason is that we will work with Hankel matrices as technical tools even ifwe do not know whether or not s is a moment sequence. In algebraic geometrythe Hankel matrix appears under the name catalecticant matrix. ı

2.3.2 Example 2: N2d0; .n;m/� D .m;n/

The additive semigroup N2d0 with involution

.n1; : : : ; nd;m1; : : : ;md/� D .m1; : : : ;md; n1; : : : ; nd/

is a �-semigroup and the map

.n1; : : : ; nd;m1; : : : ;md/ 7! zn11 � � � zndd zm11 � � � zmdd

is a �-isomorphism of CŒN2d0 � onto the �-algebra CŒz1; : : : ; zd; z1; � � � ; zd� of

complex polynomials with involution given by .zj/� WD zj, j D 1; : : : ; d. That is,

CŒN2d0 � Š CŒz1; z1; : : : ; zd; zd�:

Arguing as in the preceding example it follows that the character space is given bythe evaluations at points of Cd, that is,

S� D f�z W z 2 Cdg Š Cd; where �z. p/ D p.z1; : : : ; zd; z1; : : : ; zd/:

Positive semidefinite functions on this �-semigroup S are complex multise-quences s D .s˛;ˇ/˛;ˇ2Nd

0for which

X˛;˛0 ;ˇ;ˇ02Nd

0

s˛C˛0 ;ˇCˇ0 �˛;ˇ �˛0 ;ˇ0 � 0

for all finite complex multisequences .�˛;ˇ/˛;ˇ2Nd0.

The ..˛; ˇ/; .˛0; ˇ0//-entry of the corresponding generalized Hankel matrix H.s/is s˛C˛0 ;ˇCˇ0 . The first equality of (2.8) yields s˛C˛0 ;ˇCˇ0 D sˇCˇ0 ;˛C˛0 for


˛; ˇ; ˛0; ˇ0 2 Nd0. In the case d D 1 the matrix H.s/ has the form

H.s/ D

0BBBBBBB@

s00 s01 s02 : : : s0n : : :s01 s11 s12 : : : s1n : : :s02 s12 s22 : : : s2n : : :: : : : : : : : : : : : : : : : : :

s0n s1n s2n : : : snn : : :: : : : : : : : : : : : : : : : : :

1CCCCCCCA: (2.19)

2.3.3 Example 3: Zd;n� D �n

The additive groupZd equipped with the involution .n1; : : : ; nd/� D .�n1; : : : ;�nd/is a �-semigroup. There is a �-isomorphism

.n1; : : : ; nd/ 7! zn11 � � � zndd

of the group �-algebra CŒZd� onto the �-algebra of trigonometric polynomials in dvariables, or equivalently, of complex polynomials in z1 2 T; : : : ; zd 2 T. Thus,

CŒZd� Š CŒz1; z1; : : : ; zd; zd W z1z1 D z1z1 D 1; : : : ; zdzd D zdzd D 1�:

It is easily verified that the character space .Zd/� consists of point evaluations atpoints of the d-torus Td D f.z1; : : : ; zd/ 2 Cd W jz1j D � � � D jzdj D 1 g, that is,

.Zd/� D f�z W z 2 Tdg Š Td; where �z. p/ D p.z/:

The .n;m/-entry of the generalized Hankel matrix H.s/ is sm�n and we havesm�n D sn�m for n;m 2 Zd. In the case d D 1 the matrix H.s/ is given by

H.s/ D

0BBBBBBB@

: : : : : : : : : : : : : : : : : :

: : : s0 s1 s2 s3 : : :

: : : s1 s0 s1 s2 : : :

: : : s2 s1 s0 s1 : : :

: : : s3 s2 s1 s0 : : :

: : : : : : : : : : : : : : : : : :

1CCCCCCCA; (2.20)

where s0 stands at the (0,0)-entry. A matrix of this form is called a Toeplitz matrix.This matrix is constant on each descending diagonal from left to right.

2.4 Exercises 53

2.3.4 Example 4: Z;n� D n

The additive group Z with identity involution is a �-semigroup. The map n 7! xn isa �-isomorphism of the group �-algebra RŒZ� on the �-algebra RŒx; x�1� of Laurentpolynomials with involution given by x� D x. The character space Z� is

Z� D f�t W t 2 Rnf0gg Š Rnf0g; where �t. p/ D p.t; t�1/:

The .n;m/-entry of the generalized Hankel matrix H.s/ is snCm, so that

H.s/ D

0BBBBBBB@

: : : : : : : : : : : : : : :

: : : s�2 s�1 s0 : : :

: : : s�1 s0 s1 : : :

: : : s0 s1 s2 : : :

: : : s1 s2 s3 : : :

: : : : : : : : : : : : : : :

1CCCCCCCA: (2.21)

Here again s0 is located at the (0,0)-entry of the matrix. That is, H.s/ is a “usual”two-sided infinite Hankel matrix which is constant on cross-diagonals.

2.4 Exercises

1. Let S be an abelian �-semigroup. Show that S�S is an abelian �-semigroup withproduct .s1; s2/ ı .s0

1; s02/ D .s1s0

1; s2s02/ and involution .s1; s2/� D .s2; s1/, where

s1; s01; s2s

02 2 S. Which examples of Sect. 2.3 fit into this scheme?

2. Let s D .sn/n2N0 be a complex positive semidefinite sequence for the �-semigroup N0 with involution n� D n. Prove the following:

a. sn 2 R and s2n � 0 for n 2 N0.b. .smCn/

2 � s2ms2n for m; n 2 N0.c. .sn/2

k � .s0/2k�1sn2k for n 2 N0, k 2 N. In particular, s0 D 0 implies that

sn D 0 for all n 2 N.

3. (Schur’s theorem) Show that if A D .aij/ni;jD1 and B D .bij/ni;jD1 are positivesemidefinite matrices over R, then so is the matrix C WD .aijbij/ni;jD1.

4. Show that if s D .sn/n2N0 and t D .tn/n2N0 are positive semidefinite sequencesfor the �-semigroup N0, then so is the pointwise product sequence .sntn/n2N0 :

5. Let ' and be positive semidefinite functions on the additive group R. Showthat ', ' C , and ' are also positive semidefinite functions on R.

6. Show that '.t/ D cos t is a positive semidefinite function on R.


7. Let � 2 MC.R/ be a finite measure. Prove that

'.t/ DZR

e�i tx d�.x/; t 2 R;

is a continuous positive semidefinite function on R.(Bochner’s theorem (see e.g. [RS2]) states that each continuous positive semidef-inite function ' on R is of this form with � uniquely determined by ':)

2.5 Notes

Basics on positive functionals and general �-algebras can be found (for instance)in [Sm4]. The notion of a �-semigroup appeared first in the Appendix written byB. Sz.-Nagy of the functional analysis textbook [RzSz]. The standard monographabout harmonic analysis on semigroups is [BCRl].

Part IThe One-Dimensional Moment Problem

Chapter 3One-Dimensional Moment Problemson Intervals: Existence

In this chapter we begin the study of one-dimensional (full) moment problems:Given a real sequence s D .sn/n2N0 and closed subset K of R, the K-moment

problems asks: When does there exist a Radon measure� onR supported on K suchthat sn D

RRxnd�.x/ for all n 2 N0?

Our main aims are the solvability theorems for K D R (Hamburger’s Theo-rem 3.8), K D Œ0;C1/ (Stieltjes’ Theorem 3.12), and K D Œa; b� (Hausdorff’sTheorems 3.13 and 3.14). They are derived in Sect. 3.2 from Haviland’s theo-rem 1.12. To apply this result representations of positive polynomials in terms ofsums of squares are needed. In Sect. 3.1 we develop such descriptions that aresufficient for the applications in Sects. 3.2 and for the truncated moment problemstreated in Sects. 9.4 and 10.1.

In Sect. 3.3 we establish a one-to-one correspondence between the Stieltjesmoment problem and the symmetric Hamburger moment problem. In Sect. 3.4 wederive unique representations of nonnegative polynomials on intervals (Proposi-tions 3.20–3.22). These results are stronger than those obtained in Sect. 3.1 and theyare of interest in themselves.

3.1 Positive Polynomials on Intervals

Suppose that p.x/ 2 RŒx� is a fixed nonconstant polynomial. Since p has realcoefficients, it follows that if � is a non-real zero of p with multiplicity l, so is�. Clearly, .x � �/l.x � �/l D ..x � u/2 C v2/l, where u D Re� and v D Im�.Therefore, by the fundamental theorem of algebra, each nonzero real polynomial pfactors as

p.x/ D a.x � ˛1/n1 � � � .x � ˛r/nr .x � �1/j1 .x � �1/j1 : : : .x � �k/jk .x� �k/jk (3.1)

D a.x � ˛1/n1 � � � .x � ˛r/nr ..x � u1/2 C v21/j1 : : : ..x � uk/

2 C v2k /jk ; (3.2)


57

58 3 One-Dimensional Moment Problems on Intervals: Existence

where

n1; : : : ; nr; j1; : : : ; jk 2 N; a; ˛1; : : : ; ˛r 2 R;

�1 D u1 C iv1; : : : ; �k D uk C ivk; u1; : : : ; uk 2 R; v1 > 0; : : : ; vk > 0;

˛i ¤ ˛j if i ¤ j; and �i ¤ �j; �i ¤ �j if i ¤ j:

Thus, Eq. (3.2) expresses p as a product of a constant a, of powers of pairwisedifferent linear polynomials x � ˛i with real zeros ˛i, and of powers of pairwisedifferent quadratic polynomials .x � uj/2 C v2j with no real zeros. Note that linearfactors or quadratic factors may be absent in (3.2). Up to the numeration of factorsthe representation (3.2) (and likewise the representation (3.1)) of p is unique.

If a; b; c; d are elements of a commutative ring, there is the two square identity

.a2 C b2/.c2 C d2/ D .ac � bd/2 C .adC bc/2: (3.3)

This implies that each product of sums of two squares is again a sum of two squares.Recall that Pos.M/ is the set of p 2 RŒx� that are nonnegative on M R andPRŒx�2 is the set of finite sums of squares p2, where p 2 RŒx�. We denote by RŒx�n

and Pos.M/n the corresponding subsets of polynomials p such that deg. p/ � n andbyP2

n the set of finite sums of squares p2, where deg. p/ � n.The following three propositions contain all results on positive polynomials

needed for solving the moment problem on intervals. The formulas containingpolynomial degrees will be used later for the truncated moment problems.

Proposition 3.1

(i) Pos.R/ DPRŒx�2 D f f 2 C g2 W f ; g 2 RŒx� g.(ii) Pos.R/2n DP2

n D f f 2 C g2 W f ; g 2 RŒx�n g.Proof

(i) Let p 2 Pos.R/, p ¤ 0. Since p.x/ � 0 on R, it follows that a > 0 andthe numbers k1; : : : ; kr in (3.2) are even. Hence p is a product of squares andof sums of two squares. Therefore, by (3.3), p is of the form f 2 C g2, wheref ; g 2 RŒx�. The other inclusions are obvious.

(ii) follows at once from (i), because deg. f 2 C g2/ D 2max.deg. f /; deg.g//. utProposition 3.2

Pos.Œ0;C1// D ˚ f C xg W f ; g 2 ˙ RŒx�2 g; (3.4)

Pos.Œ0;C1//2n D˚f C xg W f 2 ˙2

n ; g 2 ˙2n�1�; n 2 N; (3.5)

Pos.Œ0;C1//2nC1 D˚f C xg W f ; g 2 ˙2

n

�; n 2 N0: (3.6)

3.1 Positive Polynomials on Intervals 59

Proof Clearly, (3.4) implies (3.5) and (3.7), since

deg.X

if 2i C xg2i / D maxi.2 deg. fi/; 1C 2 deg.gi//:

Let us abbreviate Q WDPRŒx�2C xP

RŒx�2. It is obvious that Q Pos.Œ0;C1//.Thus, it suffices to prove that Pos.Œ0;C1// Q.

Next we note that the set Q is closed under multiplication. Indeed, for arbitraryf1; f2; g1; g2 2PRŒx�2, we have

. f1 C xg1/. f2 C xg2/ D . f1f2 C x2g1g2/C x. f1g2 C g1f2/ 2 Q:

Let p 2 Pos.Œ0;C1//; p ¤ 0; and consider the representation (3.2). Since Q isclosed under multiplication, it suffices to show that all factors from (3.2) are in Q.Products of quadratic factors and even powers of linear factors are obviously in Q.It remains to handle the constant a and the linear factor x � ˛i for each real zero ˛iof odd multiplicity. Since p.x/ � 0 on Œ0;C1/, we have a > 0 by letting x!C1and ˛i � 0, because p.x/ changes its sign in the neighbourhood of a zero with oddmultiplicity. Hence a 2 Q and x�˛i D .�˛iCx/ 2PRŒx�2Cx

PRŒx�2 D Q: ut

Proposition 3.3 Suppose that a; b 2 R, a < b. Then:

Pos.Œa; b�/ D ˚ f C .x � a/g W f ; g 2 ˙ RŒx�2�; (3.7)

Pos.Œa; b�/2n D˚f C .b � x/.x � a/g W f 2 ˙2

n ; g 2 ˙2n�1�; (3.8)

Pos.Œa; b�/2nC1 D˚.b � x/f C .x � a/g W f ; g 2 ˙2

n

�: (3.9)

Proof The equality (3.7) is an immediate consequence of (3.8) and (3.9).All polynomials on the right-hand sides of (3.8) and (3.9) belong to the

corresponding left-hand sides. We prove the converse inclusions of (3.8) and (3.9)by induction on n. Both (3.8) and (3.9) are trivial for n D 0. Assume that (3.8) and(3.9) hold for n. Let p 2 Pos.Œa; b�/2nC2 or p 2 Pos.Œa; b�/2nC3.

Suppose that p has a quadratic factor q without real zeros in (3.2). Multiplying by�1 if necessary we can assume that q � 0 on R. Then p D qp0 with p0 2 Pos.Œa; b�/and deg. p0/ � deg. p/ � 2. Applying the induction hypothesis to p0 it follows thatp is in the corresponding set on the right-hand side of (3.8) or (3.9).

Now we treat the case when p has a real zero, say ˛. Upon a linear transformationwe can assume without loss of generality that a D 0 and b D 1. Then .b�x/.x�a/ Dx.1�x/. First let ˛ 2 .0; 1/. Considering p.x/ in a neighbourhood of ˛, we concludethat ˛ has even multiplicity. Hence we can factorize p D .x� ˛/2p0 and argue as inthe preceding paragraph. Thus we can assume that ˛ … .0; 1/.

Case 1: p 2 Pos.Œ0; 1�/2nC2:First suppose that ˛ � 0. Then x � ˛ � 0 on Œ0; 1�, so we can write p D .x � ˛/p0with p0 2 Pos.Œ0; 1�/2nC1. By the induction hypothesis we have

p0 2 .1 � x/f C xg with f ; g 2 ˙2n : (3.10)


Therefore, from the identity

p D .x � ˛/p0 D x.1 � x/f C x2g � ˛Œx.1 � x/. f C g/C .1 � x/2f C x2g�

we conclude that p 2P2nC1Cx.1 � x/

P2n.

Now suppose that ˛ � 1. Then p D .˛ � x/p0 with p0 2 Pos.Œa; b�/2nC1. Theassertion p 2 P2

nC1Cx.1 � x/P2

n follows from the induction hypotheses (3.10)combined with the identity

p D .˛ � x/p0 D .1 � x/p0 C .˛ � 1/p0D .1 � x/2f C x.1 � x/gC .˛ � 1/Œx.1 � x/. f C g/C .1 � x/2f C x2g�:

Case 2: p 2 Pos.Œa; b�/2nC3:First let ˛ � 0. Then we write p D .x � ˛/p0 with p0 2 Pos.Œa; b�/2nC2. Hence, bythe induction hypothesis,

p0 D f C x.1 � x/g with f 2 ˙2nC1; g 2 ˙2

n :

Then the identity

p D .x � ˛/p0 D xf C .1 � x/x2g � ˛=2 Œx .2f C .1 � x/2g/C .1 � x/.2f C x2g/�

implies that p 2 .1 � x/P2

nC1CxP2

nC1 :Now let ˛ � 1. Then it follows from

p D .˛ � x/p0 D .1 � x/p0 C .˛ � 1/p0D .1 � x/f C x.1 � x/2gC .˛ � 1/=2 Œx .2f C .1 � x/2g/C .1� x/.2f C x2g/�

that p 2 .1 � x/P2

nC1CxP2

nC1 :This completes the induction proof of (3.8) and (3.9). utAnother proof of formulas (3.8) and (3.9) is sketched in Exercise 3.2. Descrip-

tions of Pos.K/ for some other sets K are given in Exercise 3.7. In Sect. 3.4 we givestronger forms of representations of positive polynomials.

The next proposition is a classical result due to S. Bernstein. It enters into thesolution of Hausdorff’s moment problem given by Proposition 3.14 below.

Proposition 3.4 Suppose that p 2 RŒx� and p.x/ > 0 for all x 2 Œ�1; 1�. Then thereare numbers m 2 N and akl � 0 for k; l D 1; : : : ;m such that

p.x/ DmX

k;lD0akl.1 � x/k.1C x/l: (3.11)

3.1 Positive Polynomials on Intervals 61

The proof of Proposition 3.4 is based on two classical lemmas which are of interestin themselves. The following lemma is due to E. Goursat.

Lemma 3.5 Suppose that p 2 RŒx�, p ¤ 0, and m D deg. p/. The Goursattransform of p is the polynomial Qp 2 RŒx� defined by

Qp.x/ D .1C x/mp

�1 � x

1C x

�: (3.12)

Then we have:

(i) deg .Qp/ � m and we have deg.Qp/ D m if and only if p.�1/ ¤ 0.(ii) p 2 Pos.Œ�1; 1�/ if and only if Qp 2 Pos.Œ0;C1//.

(iii) p.x/ > 0 on Œ�1; 1� if and only if Qp.x/ > 0 on Œ0;C1/ and deg .Qp/ D m.

Proof

(i) Let p.x/ DPmkD0 akxk. It is obvious that

Qp.x/ DmX

kD0ak.1C x/m�k.1 � x/k

is a polynomial and deg .Qp/ � m: Its coefficient of xm isPm

kD0 ak.�1/k Dp.�1/ Thus deg.Qp/ D m if and only if p.�1/ ¤ 0.

(ii) For x ¤ �1 we set t D 1�x1Cx . Then t ¤ �1 and x D 1�t

1Ct . Further, x 2 .�1; 1�if and only if t 2 Œ0;C1/. Therefore, we have p.x/ � 0 on .�1; 1�, orequivalently p.x/ � 0 on Œ�1; 1�, if and only if Qp.t/ � 0 on Œ0;C1/.

(iii) Clearly, p.x/ > 0 for x 2 .�1; 1� if and only if Qp.t/ > 0 on Œ0;C1/: If thisholds, then p.�1/ � 0, so that p.�1/ > 0 if and only if deg .Qp/ D m by (i). ut

Remark 3.6 Let us note the following interesting facts:The inverse of the mapping x 7! t D 1�x

1Cx is given by the same formula t 7! x D 1�t1Ct .

If p and its Goursat transform Qp have degree m, then the Goursat transform of Qp isjust 2mp: ı

The next lemma is the one-dimensional version of a classical result of G. Polya;a multivariate version is given by Proposition 12.51 below.

Lemma 3.7 Suppose that p 2 RŒx� and p.x/ > 0 for x 2 Œ0;C1/. Then thereexists an N 2 N such that .1C x/Np.x/ has only positive coefficients, that is,

.1C x/Np.x/ DmX

kD0bkx

k with bk > 0; k D 0; : : : ;m:

Proof Let us introduce the notation .z/jt WD z.z � t/ : : : .z� . j� 1/t/. Then


�k

m

�j

1=m

D k

m

�k

m� 1

m

�� k

m� j � 1

m

�D kŠ

.k � j/Šmj: (3.13)

Let p.x/ DPnjD0 ajxj, where an ¤ 0, and define

P.x; y/ WDnX

jD0ajx

jyn�j and Pt.x; y/ WDnX

jD0aj.x/

jt.y/

n�jt : (3.14)

Since p.x/ > 0 on Œ0;C1/ and deg. p/ D n, the homogenous polynomial P ispositive on � D f.x; y/ W x � 0; y � 0; xC y D 1g; so P has a positive minimum,say c, on the compact set �. For N 2 N we have

.xC y/NP.x; y/ DnX

jD0

NXiD0

aj

N

i

!xiCjyNCn�i�j: (3.15)

Fix k 2 N0 such that k � m. Set m WD N C n and l WD m � k. The coefficient bk ofxkyl in (3.15) is

bk DkX

jD0aj

N

k � j

!kX

jD0aj

NŠ

.k � j/Š.N � .k � j//Š

D NŠmn

kŠ lŠ

kXjD0

ajkŠ

.k � j/Šmj

lŠ

.l � .n � j//Šmn�j

D NŠmn

kŠ lŠ

kXjD0

aj

�k

m

�j

1=m

�l

m

�n�j

1=m

D NŠmn

kŠ lŠP1=m

�k

m

��l

m

�:

Here the equality before last holds by (3.13) and the last equality is the definition ofP1=m. Since Pt.x; y/ ! P.x; y/ uniformly on � as t ! C0 and P.x; y/ � c > 0 on�, it follows that bk > 0 for all k if N, hence m D N C n, is sufficiently large. utProof of Proposition 3.4 Let n D deg. p/. Since p.x/ > 0 on Œ�1; 1�,Lemma 3.5(iii) implies that the Goursat transform Qp has degree n and Qp.x/ > 0 onŒ0;C1/. Thus, by Lemma 3.7, there are numbers N 2 N and a0 > 0; : : : ; aNCn > 0

such that

.1C t/N Qp.t/ DNCnXjD0

ajtj: (3.16)

Set m WD N C n and t D 1�x1Cx for x ¤ �1. Then x D 1�t

1Ct and .1C t/�1 D 1Cx2

, sothat Qp.t/ D .1C t/np.x/. Inserting these facts and using Eq. (3.16) we derive

3.2 The Moment Problem on Intervals 63

p.x/ D .1C t/�n Qp.t/ D .1C t/N�m Qp.t/ DNCnXjD0

ajtj.1C t/�m

DmXjD0

aj

�1 � x

1C x

�j�1C x

2

�m

DmXjD0

2�maj.1 � x/j.1C x/m�j: ut

3.2 The Moment Problem on Intervals

In this section, we solve the moment problem for closed intervals J by combiningHaviland’s theorem with the descriptions of Pos.J/ given in the preceding section.

Let P.N0/ denote the set of real sequences s D .sn/n2N0 which are positivesemidefinite, that is, for all �0; �1; : : : ; �n 2 R and n 2 N we have

nXk;lD0

skCl�k �l � 0: (3.17)

Let s D .sn/n2N0 be a real sequence. Recall that Ls is the Riesz functional onRŒx� defined by Ls.xn/ D sn, n 2 N0. Let Es denote the shifted sequence given by

.Es/n D snC1; n 2 N0:

Clearly, LEs. p.x// D Ls.xp.x// for p 2 RŒx�.Further, we define the Hankel matrix Hn.s/ and the Hankel determinant Dn.s/ by

Hn.s/ D

0BBBBB@

s0 s1 s2 : : : sns1 s2 s3 : : : snC1s2 s3 s4 : : : snC2: : : : : : : : : : : : : : :

sn snC1 snC2 : : : s2n

1CCCCCA; Dn.s/ D detHn.s/: (3.18)

The following result is Hamburger’s theorem.

Theorem 3.8 (Solution of the Hamburger Moment Problem) For a realsequence s D .sn/n2N0 the following are equivalent:

(i) s is a Hamburger moment sequence, that is, there is a measure � 2 MC.R/such that xn 2 L1.R; �/ and

sn DZR

xnd�.x/ for n 2 N0: (3.19)

(ii) s 2 P.N0/; that is, the sequence s is positive semidefinite.(iii) All Hankel matrices Hn.s/, n 2 N0, are positive semidefinite.(iv) Ls is a positive linear functional onRŒx�, that is, Ls. p2/ � 0 for p 2 RŒx�:


Proof From Proposition 2.7 we know that (i) implies (ii) and that (ii) and (iv)are equivalent. The Hankel matrix Hn.s/ is just the symmetric matrix associ-ated with the quadratic form in (3.17); hence (ii) and (iii) are equivalent. Themain implication (iv)!(i) follows from Haviland’s Theorem 1.12 combined withProposition 3.1. ut

The next proposition deals with representing measures of finite support.

Proposition 3.9 For a positive semidefinite sequence s the following are equiva-lent:

(i) There is a number n 2 N0 such that

D0.s/ > 0; : : : ;Dn�1.s/ > 0 and Dk.s/ D 0 for k � n: (3.20)

(ii) s is a moment sequence with a representing measure � supported on n points.

Proof By Theorem 3.8 the sequence s has a representing measure �. For c D.c0; c1; : : : ; ck/T 2 RkC1 we define pc.x/ WDPk

jD0 cjxj: Then, by (3.19) we derive

cTHk.s/c DkX

j;lD0sjClcjcl D

Z ˇˇ kXjD1

cjxj

ˇˇ2d�.x/ D

Zjpc.x/j2 d�.x/: (3.21)

The proof is based on the following two facts.

I. First suppose that supp� consists of n points. Then, for k � n we can choosec 2 RkC1; c ¤ 0; such that the polynomial pc.x/ vanishes on supp�. Then(3.21) is zero, so Hk.s/ is not positive definite and hence Dk.s/ D 0.

II. Suppose that Dk.s/ D 0. Then Hk.s/ is not positive definite, so there exists ac ¤ 0 such that the expression in (3.21) is zero. Therefore, by Proposition 1.23,supp� Z. pc/: Since deg. p/ � k, supp� contains at most k points.

(i)!(ii) Since Dn.s/ D 0 by (i), supp� has at most n points by II. If supp� hadfewer than n points, then we would have Dn�1.s/ D 0 by I, which contradicts (i).

(ii)!(i) Then Dk.s/ D 0 for k � n by I. If Dk.s/ were zero for some k < n � 1,then supp� would have at most n � 1 points by II. This contradicts (ii). utRemark 3.10 It was recently proved in [BS3] that the assumption “s is positivesemidefinite” in Proposition 3.9 can be omitted. That is, if s is an arbitrary realsequence satisfying condition (i), then s is a Hamburger moment sequence (and hasan n-atomic representing measure by Proposition 3.9). ı

Many considerations in subsequent chapters require the stronger assumption thatthe moment sequence s D .sn/n2N0 s is positive definite, that is,

nXk;lD0

skClckcl > 0 for all c D .c0; c1; : : : ; cn/T 2 RnC1; c ¤ 0; n 2 N0:

The following proposition characterizes this property.

3.2 The Moment Problem on Intervals 65

Proposition 3.11 For a Hamburger moment sequence s D .sn/n2N0 the followingstatements are equivalent:

(i) Each representating measure � of s has infinite support.(ii) s is positive definite.

(iii) Hn.s/ is positive definite for all n 2 N0.(iv) Dn.s/ > 0 for all n 2 N0.

Proof The equivalence (ii)$(iii) and the implication (iii)!(iv) are clear fromelementary linear algebra. Proposition 3.9 yields (i)$(iv).

It suffices to prove (i)!(iii). Assume that (3.21) vanishes for some c. Then theinfinite set supp� is contained in the zero set of the polynomial pc. Hence pc D 0,so that c D 0. Thus, Hk.s/ is positive definite for each k 2 N0. This proves (iii). ut

The second main result is Stieltjes’ theorem.

Theorem 3.12 (Solution of the Stieltjes Moment Problem) For any realsequence s the following statements are equivalent:

(i) s is a Stieltjes moment sequence, that is, there is a measure � 2 MC.Œ0;C1//such that xn 2 L1.Œ0;C1/; �/ and

sn DZ 1

0

xnd�.x/ for n 2 N0: (3.22)

(ii) s 2 P.N0/ and Es 2 P.N0/.(iii) All Hankel matrices Hn.s/ and Hn.Es/, n 2 N0, are positive semidefinite.(iv) Ls. p2/ � 0 and Ls.xq2/ � 0 for all p; q 2 RŒx�:

Proof The proof is almost the same as the proof of Theorem 3.8; instead ofProposition 3.1 we apply formula (3.5) in Proposition 3.3. ut

Combining Haviland’s theorem with (3.7) the same reasoning used in the proofsof Theorems 3.8 and 3.12 yields the following result.

Theorem 3.13 (Solution of the Moment Problem for a Compact Interval) Leta; b 2 R, a < b. For a real sequence s the following are equivalent:

(i) s is an Œa; b�-moment sequence.(ii) s 2 P.N0/ and ..aC b/Es� E.Es/� ab s/ 2 P.N0/.

(iii) Ls. p2/ � 0 and Ls..b � x/.x � a/q2/ � 0 for all p; q 2 RŒx�.

Bernstein’s theorem (Proposition 3.4) allows us to derive two solvablity criteriaof moment problems which are not based on squares of polynomials.

Theorem 3.14 Let s D .sn/n2N0 be a real sequence and let Ls be its Riesz functionalon RŒx�. Then s is a Œ�1; 1�-moment sequence if and only if

Ls..1 � x/n.1C x/k/ � 0 for all k; n 2 N0: (3.23)


Proof The only if part is obvious, since all polynomials .1 � x/n.1 C x/k arenonnegative on Œ�1; 1�. To prove the only if part we assume that condition (3.23)holds. Then, by Proposition 3.4, Ls. p/ � 0 for all strictly positive polynomials onŒ�1; 1�. Therefore, by Haviland’s theorem 1.12 (ii)!(iv), Ls is a Œ�1; 1�-momentfunctional. Hence s is a Œ�1; 1�-moment sequence. ut

Theorem 3.14 leads to the following criterion for the Hausdorff moment problem.

Theorem 3.15 A real sequence s is a Œ0; 1�-moment sequence if and only if

..I � E/ns/k �nX

jD0.�1/j

n

j

!skCj � 0 for k; n 2 N0: (3.24)

Proof By applying the bijection x 7! 12.x C 1/ of the intervals Œ�1; 1� onto Œ0; 1�

we conclude from Theorem 3.14 that s is a Œ0; 1�-moment sequence if and only ifLs..1 � x/nxk/ � 0 for k; n 2 N0: But the latter is equivalent to (3.24), since

..I � E/ns/k DnX

jD0.�1/j

n

j

!.Ejs/k D

nXjD0.�1/j

n

j

!skCj

DnX

jD0.�1/j

n

j

!Ls.x

kCj/ D Ls..1 � x/nxk/: ut

Condition (3.24) is an important property in the context of �-semigroups. Let Sbe an abelian unital semigroup with identity map as involution. For y 2 S and afunction ' on S we define the shift Ey and a mapping�y by

.Ey'/.z/ WD '.zC y/; z 2 S; and �y WD Ey � I:

A function ' W S! R is called completely monotone if '.z/ � 0 and

.�1/n.�y1 : : : �yn'/.z/ D ..I � Ey1/ : : : .I � Eyn/'/.z/ � 0 for z 2 S

and y1; : : : ; yn 2 S. Completely monotone functions are moment functions, see e.g.[BCRl, Chapter 4, Theorem 6.4]. It can be shown that condition (3.24) implies thatthe function '.n/ D sn; n 2 N0; on the semigroup S D N0 is completely monotone,so Theorem 3.15 becomes a special case of this general result.

We close this section by treating the moment problem for the �-semigroup Z

with involution n� D n. The corresponding moment problem is called the two-sidedHamburger moment problem or strong Hamburger moment problem.

Clearly, the map n 7! xn yields an isomorphism of the semigroup algebra RŒZ�

and the algebra RŒx; x�1� of real Laurent polynomials. It is easily checked that thecharacters of RŒx; x�1� are precisely the evaluations at points of R� WD Rnf0g.

3.3 The Symmetric Hamburger Moment Problem and Stieltjes Moment Problem 67

Theorem 3.16 (Solution of the Two-Sided Hamburger Moment Problem) Fora real sequence s D .sn/n2Z the following statements are equivalent:

(i) s is a moment sequence for the �-semigroup Z, that is, there exists a positiveRadon measure � onR� such that the function xn on R� is �-integrable and

sn DZR�

xn d� for all n 2 Z: (3.25)

(ii) s 2 P.Z/, that is, s is positive semidefinite on Z.(iii) Ls is a positive functional, that is, Ls. f 2/ � 0 for all f 2 RŒx; x�1�.

Proof (ii)$(iii) and (i)!(ii) follow from Proposition 2.7 and Corollary 2.16,respectively. We prove the main implication (iii)!(i).

Let p 2 RŒx; x�1�C, that is, p.x/ � 0 for all x 2 R�. Because p is a Laurentpolynomial, x2kp 2 RŒx� for some k 2 N. Since x2kp.x/ � 0 on R� and hence onR, by Proposition 3.1 there are polynomials f ; g 2 RŒx� such that x2kp D f 2 C g2.Then p D .x�kf /2 C .x�kg/2 2PRŒx; x�1�2. Hence Ls. p/ � 0 by (iii). Therefore,by Theorem 1.14,(i)!(iv), Ls is a moment functional on RŒx; x�1� Š RŒZ�, so s isa moment sequence on Z. This proves (i). ut

3.3 The Symmetric Hamburger Moment Problemand Stieltjes Moment Problem

A Radon measure � on R is called symmetric if �.M/ D �.�M/ for all Borel setsM. Let Msym

C .R/ denote the symmetric measures of MC.R/. Set RC WD Œ0;C1/.We define mappings � W R!RC and W RC!R by �.x/ D x2 and .x/ D px.

For � 2 MsymC .R/ and � 2 MC.RC/ let �C WD �.�/ 2 MC.RC/ and .�/ 2

MC.R/ denote the corresponding images of � and � under � and , respectively.That is, �C.M/ D �.��1.M// and .�/.M/ D �. �1.M//. Further, we set

�sym WD � .�/C .� /.�/�=2: (3.26)

Then we haveZ 1

0

f .y/ d�C.y/ DZR

f .x2/ d�.x/; (3.27)

ZR

g.x/ d�sym.x/ D 1

2

Z 1

0

�g.py/C g.�py/� d�.y/ (3.28)

for Borel functions f on RC and g on R if the integrals on one side exist.


Lemma 3.17 The map � 7! �C is a bijection of MsymC .R/ onto MC.RC/ with

inverse given by � 7! �sym.

Proof The proof is given by simple verifications. As samples we show that �sym issymmetric for � 2 MC.RC/ and that .�C/sym D � for � 2 MC.R/.

Let M R be a Borel set. Inserting the corresponding definitions we derive

2�sym.M/ D �. �1.M//C �..� /�1.M//D �.ft 2 RC W t2 2 Mg/C �.ft 2 RC W �t2 2 Mg/D �.ft 2 RC W �t2 2 �Mg/C �.ft 2 RC W t2 2 �Mg/ D 2�sym.�M/:

Let �C and �� denote the restrictions of � to .0;C1/ and .�1; 0/, respec-tively. Clearly, � D �.f0g/ı0 C �C C ��. We easily verify that

�.�/ D �.f0g/ı0C 2�C; .� /�.�/ D �.f0g/ı0C 2��;

so that

2.�C/sym D �.�/C .� /�.�/ D 2�.f0g/ı0 C 2�C C 2�� D 2�: ut

Proposition 3.18 Suppose that s D .sn/n2N0 is a Stieltjes moment sequence. SetOs D .Osn/n2N0 , where Os2n D sn and Os2nC1 D 0 for n 2 N0. The map � 7! �sym

is a bijection of the solutions of the Stieltjes moment problem for s on the set ofsymmetric solutions of the Hamburger moment problem for Os and the inverse of thismap is given by � 7! �C.

Proof If � solves the Hamburger moment problem for Os, then by (3.27),

Z 1

0

ynd�C.y/ DZR

x2nd�.x/ D Os2n D sn; n 2 N0;

that is, �C solves the Stieltjes moment problem for s.Conversely, let � be a solution of the Stieltjes moment problem for s. By (3.28),

ZR

xnd�sym.x/ D 1

2

Z 1

0

�.px/n C .�px/n

�d�.x/:

This number vanishes if n is odd. If n is even, say n D 2k, then it is equal tosn=2 D sk D Osn. Thus �sym is a symmetric solution of the moment problem for Os.The remaining assertions are already contained in Lemma 3.17. ut

A Hamburger (resp. Stieltjes) moment sequence is called determinate if it hasonly one representing measure in MC.R/ (resp. in MC.Œ0;C1//.Proposition 3.19 A Stieltjes moment sequence s is determinate if and only if theHamburger moment sequence Os is determinate.

3.4 Positive Polynomials on Intervals (Revisited) 69

Proof By Proposition 3.18, there is a one-to-one correspondence between solutionsof the Stieltjes moment problem for s and symmetric solutions of the Hamburgermoment problem for Os. To complete the proof it therefore suffices to show that ifthe symmetric sequence Os is indeterminate, its moment problem has at least twosymmetric (!) solutions. This will be achieved by Corollary 6.26 in Sect. 6. ut

3.4 Positive Polynomials on Intervals (Revisited)

The representation of elements of Pos.R/ as sums of two squares given inPropositions 3.1 is far from being unique. For instance, we have

x21 C x22 D .ax1 C bx2/2 C .bx1 � ax2/

2 for a; b 2 R; a2 C b2 D 1:

In this section, we develop unique representations of nonnegative polynomials onintervals. The Markov–Lukacs theorem (Corollary 3.24) enters into the proof ofTheorem 10.29 below. Except for this, these results are not used in the rest of thebook.

Throughout this section, suppose that p.x/ is a nonconstant polynomial in RŒx�.We consider the representation (3.2) and set prz.x/ WD a.x � ˛1/n1 � � � .x � ˛r/nr :

Then pnrz.x/ WD p.x/prz.x/�1 is a polynomial with leading term 1 which has no realzero. The factorization

p.x/ D prz.x/pnrz.x/ (3.29)

decomposes p into a polynomial prz.x/ which captures all real zeros of p and apolynomial pnrz.x/ which has no real zero.

Now let p 2 Pos.R/. Then the leading coefficient a of p is positive and themultiplicity of each real zero ˛j of p is even, say nj D 2kj with kj 2 N. Thus,

prz.x/ D arY

jD1.x � ˛j/2kj ; (3.30)

so that prz 2 Pos.R/ and hence pnrz 2 Pos.R/.If prz D a or pnrz D 1, then the formulas in Propositions 3.20–3.22 should be

interpreted in the obvious manner by omitting the corresponding factors.

Proposition 3.20 The polynomial p is in Pos.R/ if and only if there are integersk1; : : : ; kr 2 N and reals a > 0; c > 0, ˛1 < � � � < ˛r; x1 < x2 < � � � < x2n�1 suchthat

p.x/ D arY

jD1.x � ˛l/2kl

nYjD1.x � x2j�1/2 C c

n�1YjD1.x � x2j/

2

: (3.31)


(One of the two polynomial factors of p in (3.31) might be absent.) The numbersa; c; ˛1; : : : ; ˛r; k1; : : : ; kr; x1; : : : ; x2n�1 are uniquely determined by these require-ments.

Proof The if part is easily checked. We carry out the proof of the only if part.As noted above, the assumption p 2 Pos.R/ implies that prz has the form (3.30).The polynomial pnrz has no real zero, leading coefficient 1 and even degree, say

2n. In the case n D 0 the second main factor in (3.31) is absent. Assume now thatn 2 N. By (3.1), the polynomial pnrz has precisely n zeros, say z1; : : : ; zn, withpositive imaginary parts. Setting

f .x/ D .x � z1/ � � � .x � zn/ and g.x/ D .x � z1/ � � � .x � zn/;

we have pnr.x/ D f .x/g.x/. Then u.x/ WD 12. f .x/Cg.x// and v.x/ WD 1

2i . f .x/�g.x//are in RŒx� and satisfy

u.x/2 C v.x/2 D f .x/g.x/ D pnrz.x/: (3.32)

Now we consider the rational functions

'j.x/ D .x � zj/.x � zj/�1; j D 1; : : : ; n; and '.x/ D '1.x/ : : : 'n.x/:

Clearly, 'j is an injective map of the real line to the unit circle. Since Im zj > 0,arg'j.x/ strictly increases on .0; 2�/ as x increases on R. Therefore,

arg'.x/ D arg'1.x/C � � � C arg'n.x/

strictly increases on .0; 2�n/ as x increases on R. Hence there exist real numbersx1 < x2 < � � � < x2n�1 such that '.xk/ D .�1/k for k D 1; : : : ; 2n � 1. Since

'.x/ D f .x/

g.x/D u.x/C iv.x/

u.x/� iv.x/D u.x/2 � v.x/2 � 2iu.x/v.x/

u.x/2 C v.x/2

and u.x/; v.x/ 2 R for x 2 R, x1; x3; : : : ; x2n�1 are zeros of the real part u.x/ andx2; : : : ; x2n are zeros of the imaginary part v.x/. Since deg.u/ D n and deg.v/ Dn�1, these numbers exhaust the zero sets of u and v, respectively. The leading termof u is 1. Put c WD b2, where b is the leading term of v. Then, by (3.32),

pnrz.x/ D u.x/2 C v.x/2 DnY

jD1.x � x2j�1/2 C c


2: (3.33)

Since p D przpnrz, (3.31) follows by combining (3.30) and (3.33).To prove the uniqueness assertion we assume that a0; c0; ˛0

l ; k0l; x

0j; r

0;m0 is anotherset of numbers satisfying the above conditions. From (3.31) it is clear that p has the


leading term a D a0 and the real zeros ˛l D ˛0l with multiplicities 2kl D 2k0

l . Hencen D deg. p/� r D deg. p/� r0 D n0. Thus, it follows from (3.31) and (3.33) that

pnrz.x/ DnY

jD1.x � x0

2j�1/2 C c0n�1YjD1.x � x0

2j/2: (3.34)

Comparing (3.33) and (3.34) we obtain

q.x/ WDnY

jD1.x � x2j�1/2 �

nYjD1.x � x0

2j�1/2 D c0n�1YjD1.x � x0

2j/2 � c


2:

The proof is complete once we have shown that q.x/ � 0. Assume the contrary.Without loss of generality, let x0

1 � x1. We denote by li the number of roots ofq which are equal to xi and by ri the number of roots of q in the open interval.xi; xiC1/. Then the number of zeros of q in the interval Œx1; x2nC1� is

m WD l1 C � � � C l2nC1 C r1 C � � � C r2m�2: (3.35)

If q.x2j/ ¤ 0, then q.x2j/ > 0, and if q.x2jC1/ ¤ 0, then q.x2jC1/ < 0 by thedefinition of q. Hence, if li D liC1 D 0, there is a zero in .xi; xiC1/, so that ri � 1.Further, if li > 0, then xi is a zero of multiplicity at least 2 and so li � 2. Thepreceding implies that ri C .li C liC1/=2 � 1 for each i D 1; : : : ; 2n � 2 and hencem � l1=2C 2n� 2 by (3.35). If l1 ¤ 0, then m > 2n� 2. If l1 D 0, then x0

1 < x1 andhence q has a zero in .x0

1; x1/, since q.x01/ > 0 and q.x1/ > 0. In both cases q has at

least 2n� 1 zeros. Since deg.q/ D 2n� 2, this is the desired contradiction. utNext we consider the half-axis. Let pŒ0;C1/

rz .x/ denote the product of the constanta and all factors .x � ˛j/nj in (3.2), where ˛j 2 Œ0;C1/. Then the polynomial

pŒ0;C1/nrz .x/ WD p.x/pŒ0;C1/

rz .x/�1 has leading term 1, no zero in Œ0;C1/, and wehave

p.x/ D p Œ0;C1/rz .x/p Œ0;C1/

nrz .x/: (3.36)

Let p 2 Pos.Œ0;C1//. Then we have a > 0 and the multiplicity nl of each zero˛l 2 .0;C1/ of p is even, nl D 2kl with kl 2 N. Thus

p Œ0;C1/rz .x/ D axk0

rYlD1.x � ˛l/2kl ; (3.37)

where k0 2 N0. In particular, p Œ0;C1/rz .x/ and p Œ0;C1/

nrz .x/ are in Pos.Œ0;C1//.Proposition 3.21 Let m WD deg. p Œ0;C1/

nrz /. Then p 2 Pos.Œ0;C1// if and only ifthere are integers k0; k1; : : : ; kr 2 N, r 2 N0, and real numbers a > 0; c > 0,


0 < ˛1 < � � � < ˛r; 0 < x1 < x2 < � � � < x2n�1; (3.38)

such that

p.x/ D axk0rY

lD1.x � ˛l/2kl

nYjD1.x � x2j�1/2 C cx


2

for m D 2n;

p.x/ D axk0rY

lD1.x � ˛l/2kl

x

nYjD1.x � x2j/

2 C cnY

jD1.x � x2j�1/2

for m D 2nC 1:

These numbers are uniquely determined by p and the above requirements.

Proof It is enough to prove the only if part. For simplicity we drop the upper indexŒ0;C1/: By the formula (3.43) and the factorization (3.42) it suffices to prove thatpnrz has the form given in square brackets.

Put P.x/ WD pnrz.x2/. Since pnrz has no zero on Œ0;1/, P has no real zeros anddeg.P/ D 2m. By Proposition 3.20, P can be represented as

P.x/ DmYjD1.x � t2j�1/2 C c

m�1YjD1.x � t2j/

2; (3.39)

where t1 < t2 < � � � < t2m�1 and c > 0. Because P is even, we also have

P.x/ DmYjD1.xC t2j�1/2 C c

m�1YjD1.xC t2j/

2; (3.40)

where �t2mC1 < � � � < �t2 < �t1: Comparing (3.39) and (3.40) it follows from theuniqueness assertion of Proposition 3.20 that

t1 D �t2m�1; : : : ; tm�1 D �tmC1; tm D 0: (3.41)

Hence, setting xj WD t2mCj for j D 1; : : : ;m � 1, the inequalities in (3.38) hold.Inserting (3.41) into (3.39) we obtain

pnrz.x2/ D P.x/ D

nYjD1.x2 � x2j�1/2 C cx2

n�1YjD1.x2 � x2j/

2;

for m D 2n and

pnrz.x2/ D P.x/ D x2

nYjD1.x2 � x2j/

2 C cnY

jD1.x2 � x2j�1/2;


for m D 2nC 1. Replacing x2 by x, we obtain the formulas in square brackets. Theuniqueness assertion is easily reduced to that of Proposition 3.20. ut

Finally, we turn to the interval Œ�1; 1�. We denote by p Œ�1;1�rz .x/ the product ofjaj, all factors .x � ˛j/nj with ˛j 2 .�1; 1/, .x C 1/ni if ˛i D �1, and .1 � x/ni if

˛i D 1 in the representation (3.2) of p. As above, p Œ�1;1�nrz .x/ WD p.x/ p Œ�1;1�rz .x/�1 isa polynomial which has no zero in Œ�1; 1� and

p.x/ D p Œ�1;1�rz .x/p Œ�1;1�nrz .x/: (3.42)

Let p 2 Pos.Œ�1; 1�/. Then the multiplicity nl of all zeros ˛l 2 .�1; 1/ of p iseven, that is, nl D 2kl with kl 2 N. Thus we have

p Œ�1;1�rz .x/ D jaj.1C x/k0 .1 � x/krC1

rYlD1.x � ˛l/2kl ; (3.43)

where k0; krC1 2 N0. Further, p Œ�1;1�rz .x/ and p Œ�1;1�nrz .x/ are in Pos.Œ�1; 1�/.Proposition 3.22 Set m D deg. p Œ�1;1�nrz /. The polynomial p is in Pos.Œ�1; 1�/ if andonly if there there are numbers k0; krC1 2 N0, k1; : : : ; kr 2 N, a > 0; b > 0; c > 0,

�1 < ˛1 < � � � < ˛r < 1; � 1 < x1 < x2 < � � � < x2n�1 < 1; (3.44)

such that (3.42) and (3.43) hold and

p Œ�1;1�nrz .x/ D bnY

jD1.x � x2j�1/2 C c.1 � x2/


2; m D 2n;

p Œ�1;1�nrz .x/ D b.1C x/nY

jD1.x � x2j/

2 C c.1 � x/nY

jD1.x � x2j�1/2; m D 2nC 1:

The corresponding numbers are uniquely determined by p and these conditions.

Proof Again the if part is easily verfied. To prove the formulas for p Œ�1;1�nrz weabbreviate p D p Œ�1;1�nrz . We set t D 1Cx

1�x and define

P.t/ D p.1/�1.1C t/mp

�t � 1tC 1

�: (3.45)

(In fact, P is just the Goursat transform of the polynomial p.1/�1p.�x/.) Then wehave x D t�1

tC1 and hence

p.x/ D p.1/.1C t/�mP.t/ D p.1/

�1 � x

2

�m

P.t/: (3.46)


Note that t 2 Œ0;C1/ if and only if x 2 Œ�1; 1/. Therefore, by (3.45), p.x/ > 0 onŒ�1; 1� implies that P.t/ > 0 on Œ0;C1/. The factor p.1/�1 in (3.45) ensures thatthe polynomial P has the leading term 1. The preceding implies that P D P Œ0;C1/

nrz .Clearly, deg.P/ D m: Thus, by Proposition 3.21, P has the form

P.t/ DnY

jD1.t � t2j�1/2 C � t

n�1YjD1.t � t2j/

2 for m D 2n; (3.47)

P.t/ D tnY

jD1.t � t2j/

2 C �nY

jD1.t � t2j�1/2 for m D 2nC 1; (3.48)

where � > 0 and 0 < t1 < t2 < � � � < t2nC1. Setting xj WD tj�1tjC1 , (3.44) holds and

t � tl D 1C x

1� x� 1C xl1 � xl

D 2.x � xl/

.1 � x/.1 � xl/; l D 1; : : : ; 2nC 1:

Inserting this into (3.47) and (3.48) by using the equality t D 1Cx1�x and (3.46) we

get

p.x/ D p.1/

�1 � x

2

�m

P.t/ D bnY

jD1.x � x2j�1/2 C c.1 � x/.1C x/


2;

for m D 2n and

p.x/ D p.1/

�1 � x

2

�m

P.t/ D b.1C x/nY

jD1.x � x2j/

2 C c.1 � x/n�1YjD1.x � x2j�1/2;

for m D 2nC 1, where b; c 2 Œ0;C1/. This proves the formulas for p Œ�1;1�nrz :

The uniqueness can be shown either by repeating the corresponding reasoningfrom the proof of Proposition 3.20 or by tracing it back to the uniqueness statementin Proposition 3.21. We do not carry out the details. utRemark 3.23 Since p Œ�1;1�nrz has leading term 1, jb � cj D 1 in Proposition 3.22. ı

From the preceding Propositions 3.21 and 3.22 we easily derive nice and usefuldescriptions of positive polynomials of degree at most m D 2n resp. m D 2nC 1.

The following result is usually called the Markov–Lukacs theorem.

Corollary 3.24 For a; b 2 R, a < b, and n 2 N0, we have

Pos.Œa; b�/2n D fpn.x/2 C .b�x/.a�x/qn�1.x/2 W pn 2 RŒx�n; qn�12RŒx�n�1g;(3.49)

Pos.Œa; b�/2nC1 D f.b � x/pn.x/2 C .a� x/qn.x/

2 W pn; qn 2 RŒx�n g: (3.50)

3.5 Exercises 75

Proof The right-hand sides are obviously contained in the left-hand sides.We prove the converse inclusion. By a linear transformation we can assume that

a D �1; b D 1. Let p 2 Pos.Œ�1; 1�/m for m D 2n resp. m D 2n C 1. Collectingthe factors of p D p Œ�1;1�r p Œ�1;1�nr in the formulas of Propositions 3.22 it follows thatp belongs to the corresponding sets on the right. ut

In a similar manner Proposition 3.21 yields at once the following corollary.

Corollary 3.25 For any n 2 N0; we have

Pos.Œ0;C1//2n D fpn.x/2 C xqn�1.x/2 W pn 2 RŒx�n; qn�1 2 RŒx�n�1g; (3.51)

Pos.Œ0;C1//2nC1 D fxpn.x/2 C qn.x/2 W pn; qn 2 RŒx�n g: (3.52)

Remark 3.26 All three Propositions 3.20–3.22 give unique representations of non-negative polynomials on the corresponding intervals. Propositions 3.21 and 3.22 arestronger than Corollaries 3.25 and 3.24, while Corollaries 3.25 and 3.24 are strongerthan Propositions 3.2 and 3.3, respectively. However, as already mentioned earlier,Propositions 3.1–3.3 are sufficient for solving the moment problems on intervals. ı

3.5 Exercises

1. Let A be a commutative ring. Find the counterpart of the two square identity(3.3) for n D 4 and n D 8: Given elements a1; : : : ; an; b1; : : : ; bn 2 A, there areelements c1; : : : ; cn 2 A which are bilinear functions of the ai and bj such that

.a21 C � � � C a2n/.b21 C � � � C b2n/ D c21 C � � � C c2n:

Remark: As shown by A. Hurwitz, n D 1; 2; 4; 8 are the only natural numbersfor which there is such an n square identity, see [Hu] for precise formulation.

2. Use the Goursat transform (3.12) to derive the formulas (3.8) and (3.9) for theinterval Œ�1; 1� from the corresponding formulas (3.5) and (3.7) for Œ0;C1/.

3. Show that Pos.Œ�1; 1�/ D ff C .1 C x/g C .1 � x/h W f ; g; h 2 PRŒx�2g anduse this description to formulate a solvability criterion for the Œ�1; 1�-momentproblem.

4. Let f 2 RŒx�, a; b 2 R; a < b, and set Tf D fpC fq W p; q 2PRŒx�2g. Supposethat Œa; b� D fx 2 R W f .x/ � 0g. Let k and l be the multiplicities of the zeros aand b of f , respectively.

a. Find a polynomial q 2 Pos.Œ�1; 1�/ such that q … Tf for f .x/ D .1 � x2/3.b. Show that Pos.Œa; b�/ D Tf if and only if k D l D 1.

Details for b. can be found in the proof of [PR, Corollary 11].5. Let p.x/ D x2 C c, where 0 < c < 1. Show that p cannot be written in the form

p.x/ DP2k;lD0 akl.1 � x/k.1C x/l with akl � 0 for k; l D 0; 1; 2.


6. Let a; b; c 2 R, a < b < c, and set K D Œa; b� [ fcg. Describe Pos.K/ and findsolvability conditions for the K-moment problem.

7. Let a; b; c; d 2 R; a < b < c < d. Show that

a. Pos..�1; a�[ Œb;C1// D ˚ f C .x � a/.x � b/g W f ; g 2PRŒx�2�;

b. Pos.Œa; b�[ Œc; d�/ D ˚ f C .x�a/.b�x/.x�c/.x�d/g W f ; g 2PRŒx�2�:

8. Use Exercise 7 to give solvability criteria for the K-moment problem, where

a. K D .�1; a�[ Œb;C1/,b. K D Œa; b� [ Œc; d�.

9. Suppose that �1 < a1 < b1 < a2 < � � � < an < bn < C1. Define K1 WD[n

kD1Œak; bk� and K2 WD Rn [nkD1 .ak; bk/: Describe Pos.Kj/ and give necessary

and sufficient conditions for Kj-moment sequences, where j D 1; 2.10. Show that the sequence s D .0; 1; 0; 0; : : : / satisfies Dk.s/ D 0 for all k 2 N0,

but s is not a moment sequence.11. Give an alternative proof of Proposition 3.4 by showing following steps:

a. It suffices to prove the result for linear and for quadratic polynomials.b. The assertion holds for linear and for quadratic polynomials.

(This proof was given by F. Hausdorff [Hs], p. 98–99, see e.g [PSz], p. 276–277.)

12. Suppose that s is a moment sequence. Prove thatP2n

kD0skkŠ � 0 for n 2 N.

13. Let K be a closed subset of R. Prove that the following statements areequivalent:

(i) If s D .sn/ and t D .tn/ are K-moment sequences, then so is st WD .sntn/.(ii) If x; y 2 K, then xy 2 K.

Hint: For (ii))(i), use Haviland’s theorem and Lst;z. f .z// D Ls;x�Lt;y. f .xy//

�:

14. “Guess” representing measures for the following sequences .sn/1nD0:

a. sn D anC1

nC1 C cbn, where b 2 R and a; c � 0,b. sn D nŠ;c. sn D 1

.nC1/.nC2/ :

15. (Solution of the two-sided Stieltjes moment problem)Show that for a real sequence s D .sn/n2Z the following are equivalent:

(i) s is a two-sided Stieltjes moment sequence, that is, there exists a Radonmeasure � on .0;C1/ such that xn is �-integrable and sn D

R C10 xn d�

for n 2 Z.(ii) s 2 P.Z/ and Es 2 P.Z/.

(iii) Ls and LEs are positive functionals on RŒx; x�1�, that is, Ls. f 2/ � 0 andLs.xf 2/ � 0 for all f 2 RŒx; x�1�.

3.6 Notes 77

3.6 Notes

The results on positive polynomials have a long and tricky history, see [PR] forsome discussion. Polya’s theorem appeared in [P]; we reproduced his proof. TheMarkov–Lukacs theorem is due to A.A. Markov [Mv1] for m D 2n and F. Lukacs[Lu]. Proposition 3.22 is due to S. Karlin and L.S. Shapley [KSh, p. 35], whileProposition 3.21 can be found in [KSt, p. 169]. The proof of Proposition 3.20follows [Ml]; our proof of Proposition 3.22 seems to be new. Other proofs of theMarkov–Lukacs theorem are given in [Sz],[KSt] (see also [Ka]) and [KN]; Szegö[Sz, p. 4] derived it from the Fejér–Riesz theorem. Bernstein’s theorem was provedin [Bn].

The existence criteria for moment problems on intervals were obtained in theclassical papers by T.J. Stieltjes [Stj], H. Hamburger [Hm] and F. Hausdorff [Hs].The results on symmetric Hamburger moment problems are from [Chi2].

Chapter 4One-Dimensional Moment Problems:Determinacy

The main aim of this chapter is to develop some very useful results concerningthe uniqueness of solutions of one-dimensional Hamburger and Stieltjes momentproblems. These are Carleman’s conditions (4.2) and (4.3) in Theorem 4.3, whichare sufficient for determinacy, and Krein’s conditions (4.19) and (4.23) in Theo-rems 4.14 and 4.17, which provide necessary criteria.

4.1 Measures Supported on Bounded Intervals

The following proposition contains a number of characterizations of measuressupported on an interval Œ�c; c�, c > 0, in terms of their moment sequences.

Proposition 4.1 Suppose that s D .sn/n2N0 is a Hamburger moment sequence.Let � be representing measure of s and let c 2 R; c > 0. The following areequivalent:

(i) � is supported on Œ�c; c�.(ii) There exists a number d > 0 such that jsnj � dcn for n 2 N0.

(iii) There exists a number d > 0 such that s2n � dc2n for 2 N0.

(iv) S WD lim infn!1 s12n2n � c.

Further, if s0 D �.R/ � 1, then the following statements are equivalent:(v) � is supported on Œ�1; 1�.

(vi) lim infn!1 s2n � 1.(vii) lim infn!1 s2n < C1.

Proof The implications (i)!(ii)!(iii)!(iv) are obviously true with d D s0.Therefore, for the equivalence of (i)–(iv) it suffices to prove that (iv) implies (i).


79

80 4 One-Dimensional Moment Problems: Determinacy

For ˛ > 0, let �˛ denote the characteristic function of the set Rn.�˛; ˛/ and putM˛ WD

R�˛d�. Then

M˛˛2n D

ZR

˛2n�˛d� �ZR

x2n�˛d� �ZR

x2nd� D s2n

and hence .M˛/12n˛ � s

12n2n for n 2 N. Therefore, if M˛ > 0, by passing to the limits

we obtain ˛ � S. Thus, M˛ D �.Rn.�˛; ˛// D 0 when ˛ > S. Since S � c by(iv), this implies that supp� Œ�S; S� Œ�c; c�, which proves (i).

We verify the equivalence of (v)–(vii). Since �.R/ � 1, (v) implies s2n � 1 andhence (vi). The implication (vi)!(vii) is trivial, so it remains to prove (vii)!(v).

Assume to the contrary that (v) does not hold. Then we can find an intervalŒa; b� RnŒ�1; 1� such that �.Œa; b�/ > 0: Set A D a if a > 1 and A D �b ifb < �1. Then s2n � A2n�.Œa; b�/ for n 2 N. Since A > 1 and �.Œa; b�/ > 0, wededuce that limn s2n D C1: This contradicts (vii). utCorollary 4.2 If a Hamburger moment sequence s has a representing measure withcompact support, then s is determinate.

Proof Let �1; �2 2Ms. By Proposition 4.1 (iv)!(i), �1 and �2 are supported onŒ�S; S�. Then, for all f 2 RŒx�,

Z S

�Sf .x/ d�1 D

Z S

�Sf .x/ d�2: (4.1)

Since the polynomials RŒx� are dense in C.Œ�S; S�IR/ by the Weierstrass theorem,(4.1) holds for all continuous functions f . This in turn implies that �1 D �2: ut

4.2 Carleman’s Condition

Recall that a Hamburger moment sequence is determinate if it has a uniquerepresenting measure, while a Stieltjes moment sequence is called determinate ifit has only one representing measure supported on Œ0;C1/.

The Carleman theorem contains a powerful sufficient condition for determinacy.

Theorem 4.3 Suppose that s D .sn/n2N0 is a positive semidefinite sequence.

(i) If s satisfies the Carleman condition

1XnD1

s� 12n

2n D C1; (4.2)

then s is a determinate Hamburger moment sequence.

4.2 Carleman’s Condition 81

(ii) If in addition Es D .snC1/n2N0 is positive semidefinite and

1XnD1

s� 12n

n D C1; (4.3)

then s is a determinate Stieltjes moment sequenc..

The main technical ingredient of the proof of Theorem 4.3 given in this section isa result on quasi-analytic functions (Corollary 4.5). Another proof of Theorem 4.3based on Jacobi operators can be found at the end of Sect. 6.4.

Let us begin with some notions on quasi-analytic functions. Suppose .mn/n2N0

is a positive sequence and J R is an open interval. Let Cfmng denote the set offunctions f 2 C1.J/ for which there exists a constant Kf > 0 such that

supt2Jj f .n/.t/j � Kn

f mn for n 2 N0: (4.4)

We say Cfmng is a quasi-analytic class if the following holds: if f 2 Cfmng andthere is a point t0 2 J such that f .n/.t0/ D 0 for all n 2 N0, then f .t/ � 0 on J. Inthis case the functions of Cfmng are called quasi-analytic.

Quasi-analyticity is characterized by the following Denjoy–Carleman theorem.

Theorem 4.4 Cfmng is a quasi-analytic class if and only if1XnD1

.infk�n m1=kk /�1 D1: (4.5)

Proof [Hr, Theorem 1.3.8]. For log convex sequences .mn/n2N0 a proof is containedin [Ru2, Theorem 19.11]. ut

For our proof of Theorem 4.3 the following corollary is sufficient.

Corollary 4.5 Suppose that .mn/n2N0 is a positive sequence such that

1XnD1

m�1=nn D1: (4.6)

Suppose that f 2 C1.J/ and there is a constant Kf > 0 such that (4.4) is satisfied.If there exists a t0 2 J such that f .n/.t0/ D 0 for all n 2 N0, then f .t/ � 0 on J.

Proof Since obviously m1=nn � infk�n m1=kk , (4.6) implies (4.5). Hence Cfmng is aquasi-analytic class by Theorem 4.4. This proves the assertion. ut

The simplest examples of quasi-analytic functions are analytic functions.

Example 4.6 (.mn D nŠ/n2N0) Since nŠ � nn, the sequence .nŠ/n2N0 satisfies (4.5)and (4.6). Hence CfnŠg is a quasi-analytic class. It is well-known (see e.g. [Ru2,Theorem 19.9]) that each function f 2 CfnŠg has a holomorphic extension to a strip


fz W Re z 2 J; jIm zj < ıg, ı > 0. Therefore, if f 2 CfnŠg and f .n/.t0/ D 0 for somet0 2 J and all n 2 N0, then we have f .t/ � 0 on J. That is, for the special classCfnŠg the assertion of Corollary 4.5 holds by the uniqueness theorem for analyticfunctions without refering to the Denjoy–Carleman theorem. ıRemark 4.7

1. Let .mn/n2N0 be a positive sequence such that m0 D 1 and

m2n � mn�1mnC1 for n 2 N: (4.7)

Condition (4.7) implies that .ln mn/n2N0 is a convex sequence. Indeed, it can beshown that then m1=nn � m1=kk for n � k, so that m1=nn D infk�n m

1=kk . Therefore, by

Theorem 4.4, in this case Cfmng is quasi-analytic if and only ifP1

nD1m�1=nn D

1, that is, if (4.6) is satisfied.2. Let s D .sn/n2N0 be a Hamburger moment sequence such that s0 D 1. Then the

Hölder inequality implies that (4.7) holds for mn D s2n. Therefore, since m0 D 1,it follows from the preceding remark that .ln s2n/n2N0 is a convex sequence. ıThe following simple lemma is used in the proofs of Theorems 4.3 and 4.14.

Let MC.R/ denote the Radon measures � on R satisfyingR jxjn d� < 1 for all

n 2 N0: Recall that Ms is the set of representing measures of a moment sequence s.

Lemma 4.8 Suppose that � 2MC.R/ and � 2 L1.R; �/. Then the function

g.t/ WDZR

eitx�.x/d�.x/ (4.8)

is in C1.R/ and satisfies

g.n/.t/ DZR

.ix/neitx�.x/d�.x/ for n 2 N0; t 2 R: (4.9)

Proof We proceed by induction on n. For n D 0 the assertion holds by definition.Suppose that (4.9) is valid for n and all t 2 R. Fix t 2 R and put

'h.x/ WD h�1.eihx � 1/; h.x/ WD 'h.x/.ix/neitx�.x/ for h 2 R; h ¤ 0:

Then h.x/! .ix/nC1eitx�.x/ on R as h! 0. By the complex version of the meanvalue theorem, j'h.x/j D j'h.x/ � 'h.0/j � jxj sup f j' 0

h. y/j W jyj � jxj; y 2 Rg:Therefore, since j' 0

h. y/j D jeihyj D 1, we get

j h.x/j D j'h.x/.ix/neitx�.x/j � jxnC1j k�kL1.R;�/ a:e: on R:

Therefore, since � 2MC.R/ and hence xnC1 2 L1.R; �/, Lebesgue’s dominatedconvergence theorem applies and yields

4.2 Carleman’s Condition 83

limh!0

g.n/.tC h/� g.n/.t/

hD lim

h!0

ZR

h.x/d�.x/ DZR

.ix/nC1eitx�.x/d�.x/;

which gives (4.9) for nC 1. utAn immediate consequence of Lemma 4.8 is the following

Corollary 4.9 Let � 2 MC.R/. Then the Fourier transform f�.t/ WDRRe�itxd�.x/ of � is in C1.R/ and

f .n/� .t/ DZR

.�ix/neitxd�.x/ for n 2 N0; t 2 R: (4.10)

In particular,

sn.�/ DZR

xnd� D .�i/nf�.0/ for n 2 N0: (4.11)

Proof of Theorem 4.3 By Hamburger’s theorem 3.8 and Stieltjes’ theorem 3.12 itsuffices to prove the the assertions about the determinacy.

(i) Suppose that �1; �2 2 Ms and set f WD f�1 � f�2 . Then f 2 C1.R/ byCorollary 4.9. Define mn D supt2R j f .n/.t/j for n 2 N0. By (4.10),

m2n � supt2R

.j f .2n/�1.t/j C j f .2n/�2

.t/j/ �ZR

x2nd�1.x/CZ

x2nd�2.x/ D 2s2n

for n 2 N. Hence condition (4.6) is fulfilled, since

1XnD1

m�1=nn �

1XnD1

m�1=2n2n �

1XnD1

2�1=2ns�1=2n2n � 1

2

1XnD1

s�1=2n2n D1:

Applying again (4.10) we obtain for n 2 N0,

f .n/.0/ D f .n/�1.0/� f .n/�2

.0/ DZ.ix/nd�1.x/�

Z.ix/nd�2.x/ D insn � insn D 0:

Thus the assumptions of Corollary 4.5 are satisfied. Therefore, f .t/ � 0, hencef�1.t/ � f�2.t/, on R. Since the Fourier transform uniquely determines a finitemeasure, we get �1 D �2. This shows that s is determinate.

(ii) Let �1 and �2 be two solutions of the Stieltjes moment problem for s.Define symmetric measures �j D �

symj ; j D 1; 2, on R by (3.26). Then, by

Proposition 3.18, �j 2 MC.R/ has the moment sequence Os D .Osn/n2N0 ,where Os2n D sn and Os2nC1 D 0 for n 2 N0. In particular, �1 and �2 arerepresenting measures of Os. Since Os2n D sn, it follows from assumption (4.3)


that Os satisfies (4.2). Hence, by (i), the moment problem for Os is determinate.Therefore, �1 D �2 and hence �1 D �2. ut

Corollary 4.10 Suppose that s D .sn/n2N0 is a positive semidefinite sequence.

(i) If there is a constant M > 0 such that

s2n � Mn.2n/Š for n 2 N; (4.12)

then Carleman’s condition (4.2) holds and s is a determinate moment sequence.(ii) If Es D .snC1/n2N0 is also positive semidefinite and there is an M > 0 such that

sn � Mn.2n/Š for n 2 N; (4.13)

then s is a determinate Stieltjes moment sequence.

Proof

(i) For n 2 N we have .2n/Š � .2n/2n. It follows that Œ.2n/Š�1=2n � 2n and hence12n � Œ.2n/Š��1=2n, so that

M�1=2 12n� M�1=2Œ.2n/Š��1=2n � s�1=2n

2n ; n 2 N:

Therefore, Carleman’s condition (4.2) is satisfied, so that Theorem 4.3(i)applies. (As noted in Example 4.6, in this case the Denjoy–Carleman theoremis not needed.)

(ii) follows from Theorem 4.3(ii) by the same reasoning. utCorollary 4.11

(i) Let � 2 MC.R/. If there exists an " > 0 such thatZR

e"jxj d�.x/ <1; (4.14)

then� 2MC.R/, condition (4.12) holds, and the Hamburger moment problemfor � determinate.

(ii) Suppose that � 2 MC.Œ0;C1//. If there exists an " > 0 such thatZR

e"pjxj d�.x/ <1; (4.15)

then � 2MC.Œ0;C1//, condition (4.13) is satisfied, and the Stieltjes momentproblem for � is determinate.

4.3 Krein’s Condition 85

Proof

(i) Let n 2 N0 and x 2 R: Clearly, we have e"jxj � ."x/2n 1.2n/Š . Hence

x2ne�"jxj � "�2n.2n/Š (4.16)

and thereforeZR

x2nd�.x/ DZR

x2ne�"jxje"jxjd�.x/ � "�2n.2n/ŠZR

e"jxjd�.x/: (4.17)

Thus, by the assumption (4.14), we haveRx2nd�.x/ < 1 for n 2 N0 and

henceR jxkjd�.x/ < 1 for all k 2 N0, so that � 2 MC.R/: Further, (4.17)

implies that s2n � Mn.2n/Š for n 2 N0 and some constant M > 0. Thus (4.12)is satisfied and the assertion follows from Corollary 4.10(i).

(ii) follows in a similar manner with x 2 Œ0;C1/ and using Corollary 4.10(ii). utIn probability theory the sufficient determinacy conditions (4.14) and (4.15) are

called Cramer’s condition and Hardy’s condition, respectively.The examples treated below indicate that Carleman’s condition (4.2) is an

extremely powerful sufficient condition for determinacy. Nevertheless, this condi-tion is not necessary, as shown by Example 4.18 and also by Remark 7.19.

Remark 4.12 L.B. Klebanov and S.T. Mkrtchyan [KlM] proved the following:

Let s D .sn/n2N0 ; s0 D 1, be a Hamburger moment sequence. Set Cm WDPmnD1 s

� 12n

n .If � and � are representing measures of s, then

L.�; �/ � c.s2/ C�1=4m log.1C Cm/ for m 2 N; (4.18)

where c.s2/ > 0 is a constant depending only on s2 and L.�; �/ denotes the Levydistance of � and � (see e.g. [Bl]).

If Carleman’s condition (4.2) holds, then limm!1 Cm D C1, hence (4.18)implies L.�; �/ D 0, so that � D �. This is another proof of Carleman’sTheorem 4.3(i). ıRemark 4.13 C. Berg and J.P.R. Christensen [BC1] proved that Carleman’s con-dition (4.2) implies the denseness of CŒx� in Lp.R; �/ for p 2 Œ1;C1/; where� is the unique representing measure of s. We sketch a proof of this result inExercise 4.7. ı

4.3 Krein’s Condition

The following theorem of Krein shows that, for measures given by a density, theso-called Krein condition (4.19) is a sufficient condition for indeterminacy.

Theorem 4.14 Let f be a nonnegative Borel function on R. Suppose that themeasure � defined by d� D f .x/dx is in MC.R/, that is, � has finite moments


sn WDRxnd�.x/ for all n 2 N0. If

ZR

ln f .x/

1C x2dx > �1; (4.19)

then the moment sequence s D .sn/n2N0 is indeterminate.Moreover, the polynomialsCŒx� are not dense in L2.R; �/.

Proof The proof makes essential use of some fundamental results on boundaryvalues of analytic functions in the upper half plane. (All facts needed for this proofcan be found e.g. in [Gr]). Recall that the Hardy space H1.R/ consists of all analyticfunctions h in the upper half-plane satisfying

supy>0

ZR

jh.xC iy/jdx <1:

Each h.z/ 2 H1.R/ has a nontangential limit function h.x/ 2 L1.R/ [Gr, Theorem3.1]. From assumption (4.19) it follows that there is an h 2 H1.R/ such that jh.x/j Df .x/ a.e. on R [Gr, Theorem 4.4]. (In fact, (4.19) implies that the Poisson integral

u.z/ D u.xC iy/ D 1

�

ZR

y

.x � t/2 C y2ln f .t/ dt

of the function ln f .x/ exists. The corresponding function is h.z/ D eu.z/Civ.z/, wherev is harmonic conjugate to u.) Since h 2 H1.R/, it follows from a theorem of Paley–Wiener [Gr, Lemma 3.7 or p. 84] that

ZR

eitxh.x/dx D 0 for t � 0: (4.20)

Set �.x/ D h.x/f .x/�1 if f .x/ ¤ 0 and �.x/ D 0 if f .x/ D 0. Then j�.x/j � 1 onR and d� D fdx, so Lemma 4.8 applies to the function

g.t/ WDZR

eitxh.x/dx DZR

eitx�.x/d�.x/:

Recall that g.t/ D 0 for t � 0 by (4.20). Therefore, by formula (4.9) in Lemma 4.8,

.�i/ng.n/.0/ DZR

xn�.x/d�.x/ DZR

xnh.x/dx D 0 for n 2 N0: (4.21)

Let h1.x/ WD Re h.x/ and h2.x/ WD Im h.x/. From (4.21) we obtain

ZR

xnh1.x/dx DZR

xnh2.x/dx D 0 for n 2 N0: (4.22)


Since f ¤ 0 by (4.19) and jh.x/j D f .x/ a.e. on R, h1 or h2 is nonzero, say hj, andwe have f .x/ � hj.x/ � 0 on R. Hence the positive Radon measure � on R givenby d� WD . f .x/ � hj.x//dx has the same moments as � by (4.22). But � is differentfrom �, because hj ¤ 0.

Since h is nonzero, so is �. By (4.21), � 2 L2.R; �/ is orthogonal to CŒx� inL2.R; �/. Hence CŒx� is not dense in L2.R; �/. ut

Let us briefly discuss the Krein condition (4.19). First we note that (4.19) impliesthat f .x/ > 0 a.e. (Indeed, if f .x/ D 0, hence ln f .x/ D �1, on a set with nonzeroLebesgue measure, then the integral in (4.19) is �1.)

We set lnCx WD max .ln x; 0/ and ln�x WD �min .ln x; 0/ for x � 0. Thenln�x � 0 and ln x D lnCx � ln�x: Since f .x/ � 0 and hence lnC f .x/ � f .x/, wehave

0 �ZR

lnC f .x/

1C x2dx �

ZR

f .x/

1C x2dx �

ZR

f .x/ dx D s0 < C1:

Therefore, (4.19) is equivalent to

ZR

ln� f .x/

1C x2dx < C1:

Remark 4.15 In the literature, the integral

1

�

ZR

ln f .x/

1C x2dx

is often called the entropy integral or logarithmic integral, see [Ks]. ıRemark 4.16 An interesting converse of the preceding theorem was proved by J.-P. Gabardo [Gb]. Suppose that s is an indeterminate moment sequence. Then thereexists a solution of the moment problem for s given by a density f .x/ such that(4.19) holds and the entropy integral is maximal among all densities of absolutelycontinuous solutions of the moment problem. ı

The next theorem is about Krein’s condition for the Stieltjes moment problem.

Theorem 4.17 Let f , �, and s be as in Theorem 4.14. If the measure � 2MC.R/is supported on Œ0;C1/ and

ZR

ln f .x2/

1C x2dx �

Z 1

0

ln f .x/

.1C x/

dxpx> �1; (4.23)

then the Stieltjes moment problem for s is indeterminate.


Proof Note that d� D f . y/dy and � 2 MC.RC/. We define a symmetric measure� 2 Msym

C .R/ by d� D jxj f .x2/dx and we compute for h 2 Cc.RC/,Z 1

0

h. y/ d�. y/ D 2Z 1

0

h.x2/xf .x2/ dx DZR

h.x2/jxj f .x2/ dx DZR

h.x2/ d�.x/:

Comparing this equality with (3.27) we conclude that � D �sym, that is, �sym hasthe density jxj f .x2/. By assumption (4.23), we have

ZR

ln jxj f .x2/1C x2

dx DZR

ln jxj1C x2

dxCZ 1

0

ln f . y/

.1C y/

dypy> �1:

Therefore, by Theorem 4.14, the Hamburger moment sequence of�sym is indetermi-nate, so the Stieltjes moment sequence s is indeterminate by Proposition 3.19. ut

We apply the preceding criteria to treat a number of examples.

Example 4.18 (The Hamburger moment problem for d� D e�jxj˛dx, ˛ > 0)Clearly, � 2 MC.R/. If 0 < ˛ < 1, Krein’s condition (4.19) is satisfied,

since

ZR

ln e�jxj˛

1C x2dx D

ZR

�jxj˛1C x2

dx > �1:

Therefore, the Hamburger moment problem for � is indeterminate.If ˛ � 1, then

sn D�Z 1

�1CZ

jxj�1

�xne�jxj˛dx � 2C

ZR

xne�jxjdx D 2C 2nŠ � 2nnŠ: (4.24)

Hence, by Corollary 4.10(i), the Hamburger moment problem for � is determinate.ı

Example 4.19 (The Stieltjes moment problem for d� D �Œ0;1/.x/e�jxj˛dx, ˛ > 0)If 0 < ˛ < 1=2, then (4.23) holds, so the Stieltjes moment problem is

indeterminate. If ˛ � 1=2, then 2˛ � 1 and hence by (4.24),

sn DZ 1

0

xnd�.x/ DZ 1

0

.x2/nd�.x2/ DZ 1

0

x2ne�jxj2˛dx � 4n.2n/Š :

Thus, by Corollary 4.10(ii), the Stieltjes moment problem is determinate. Since �has no atom at 0, it follows from Corollary 8.9 proved in Chap. 8 that the Hamburgermoment problem for s is also determinate if ˛ � 1=2.

By some computations it can be shown that the moments of � are

sn D � ..nC 1/=˛/ for n 2 N0:


The Gamma function has an asymptotics � .x/ � p2�e�xxxC1=2 as jxj ! 1, seee.g. [RW, p. 279]. From this it follows that we have an asymptotics s1=nn � cn1=˛ for

some c > 0. HenceP1

nD1 s� 12n

2n <1 for 0 < ˛ < 1.Therefore, by the preceding, if 1=2 � ˛ < 1, then the Hamburger moment

sequence s is determinate, but Carleman’s condition (4.2) is not satisfied! Anotherexample of this kind can be found in Remark 7.19 in Sect. 7. ıExample 4.20 d� D 1

2e�jxjdx and d�.x/ D 1

2�Œ0;C1/.x/ e�p

xdx.The moments of the measures � and � are easily computed. They are

sn D 1

2

ZR

xne�jxjdx D .2n/Š ; tn D 1

2

Z 1

0

xne�pxdx D .2nC 1/Š ; n 2 N0;

so the sequences s and t satisfy conditions (4.12) and (4.13), respectively. Therefore,by Corollary 4.10 (i) and (ii), the Hamburger moment problem for � is determinateand the Stieltjes moment problem for � is determinate. Note that in both cases thecorresponding Krein conditions (4.19) and (4.23) are violated. ıExample 4.21 d�.x/ D �.0;C1/.x/x˛e�x2dx, where ˛ > �1.

In this case we calculate

sn DZ 1

0

xnx˛e�x2dx D � ..nC ˛ C 1/=2/; n 2 N0: (4.25)

Then Corollary 4.11 applies and implies that s is determinate. ıExample 4.22 (Lognormal distribution) The first examples of indeterminate mea-sures were given by T. Stieltjes in his memoir [Stj]. He showed that the log-normaldistribution d� D f .x/dx with density

f .x/ D 1p2�

�.0;C1/.x/x�1exp.�.ln x/2=2/

is indeterminate. We carry out his famous classical example in detail.Let n 2 Z. Substituting y D ln x and t D y � n, we compute

sn DZR

xn d�.x/ D 1p2�

Z 1

0

xn�1e�.ln x/2=2 dx D 1p2�

ZR

enye�y2=2 dy

D 1p2�

ZR

e�. y�n/2=2en2=2 dy D en

2=2 1p2�

ZR

et2=2 dt D en

2=2;

so that s D .en2=2/n2N0 . This proves � that finite moments, that is, � 2MC.R/.For arbitrary c 2 Œ�1; 1� we define a positive (!) measure �c by

d�c.x/ D Œ1C c sin .2� ln x/� d�.x/:


Clearly,�c 2MC.R/; since � 2MC.R/: For n 2 Z, a similar computation yields

ZR

xnsin .2� ln x/ d�.x/ D 1p2�

ZR

eny .sin 2�y/ e�y2=2dy

D 1p2�

ZR

e�. y�n/2=2en2=2sin 2�y dy D 1p

2�en2=2

ZR

e�t2=2sin 2�.tC n/ dt D 0;

where we used that the function sin 2�.tC n/ is odd. From the definition of �c andthe preceding equality it follows at once that �c has the same moments as �: (Thiswas even shown for all moments sn with n 2 Z.) Hence � is indeterminate. ıExample 4.23 (Lognormal distributions (continued)) Let ˛ 2 R and r > 0 bearbitrary. Then the function

f .x/ D 1

rp2�

�.0;1/.x/ x�1 e� .lnx�˛/2

2r2

defines a probability measure � on R by d� D f .x/dx: This measure � is

indeterminate and it has the moments sn D en˛Cn2 r22 for n 2 N0: ı

4.4 Exercises

1. Show that each Stieltjes moment sequence s D .sn/ satisfying

1XnD1

s� 12n

n D C1

is determinate as a Hamburger moment sequence.2. Suppose that .an/n2N0 is a sequence of positive numbers such that

2 ln a2n � ln anC1 C ln an�1 for n 2 N:

a. Show that npan nC1pa0 � nC1

panC1 npa0 for n 2 N0:

b. Suppose that a0 D 1. Show that . npan/n2N0 is monotone increasing.

3. Show that for a moment sequence s D .sn/ the following are equivalent:

(i) s satisfies Carleman’s condition (4.2).(ii)

P1nD1 s

�1=.4n/4n D1:

(iii)P1

nD1 s�1=.4nC2/4nC2 D 1:

(iv)P1

nD1 s�1=.2knC2l/2knC2l D 1 for some (and then for all) k 2 N; l 2 N0.

4. Prove that the moment sequence (4.25) in Example 4.21 is determinate.

4.5 Notes 91

5. Prove that the measure in Example 4.23 is indeterminate.6. Let� 2 MC.R/. Suppose that

RRx2ne"x

2d� < C1 for some " > 0 and n 2 N0.

Show that � 2MC.R/ and the moment sequence of � is determinate.7. (Carleman’s condition implies denseness of CŒx� in Lp.R; �/; p 2 Œ1;C1/

[BC2])Let � 2 MC.R/. Suppose the moment sequence of � satisfies Carleman’scondition (4.2). Prove that CŒx� is dense in Lp.R; �/ for any 1 � p < C1.

Sketch of proof It suffices to prove the denseness for p D 2k; k 2 N: Wemimic the proof of Lemma 4.8. Let � 2 Lp.R; �/0 Š Lq.R; �/, 1

p C 1q D 1;

be such thatR�.x/f .x/d� D 0 for all f 2 CŒx�. Define g.t/ by (4.8). As in the

proof of Lemma 4.8, we show that g 2 C1.R/ and that Eq. (4.9) holds. SinceR�.x/f .x/d� D 0 for f 2 CŒx�, g.n/.0/ D 0 for n 2 N0 by (4.9). Applying the

Hölder inequality to (4.9) we obtain jg.n/.t/j � s�1=.2k/2kn k�kLq.R;�/. By Exercise

4.3, (4.2) impliesP1

nD1 s�1=.2kn/2kn D C1. Thus, Corollary 4.5 applies with

mn D s1=.2k/2kn , t0 D 0; and yields g.t/ � 0. Hence � D 0, so CŒx� is dense inLp.R; �/.

8. (Moment generating function)Let � 2 MC.R/ and c > 0. Suppose that the function x 7! etx is �-integrablefor jtj < c. Then g.t/ WD R

Retxd�.x/ is called the moment generating function

of �.

a. Show that � 2 MC.R/; that is, � has finite moments sn DRxnd� for

n 2 N0.b. Show that sn D g.n/.0/ for n 2 N0.c. Show that g.t/ DP1

nD0 tn

nŠ sn for t 2 .�c; c/:

4.5 Notes

Carleman’s condition and Theorem 4.3(i) are due to T. Carleman [Cl].Theorem 4.14 is stated in [Ak, Exercise 14 on p. 87] where it is attributed to

M.G. Krein. It follows from Krein’s results in [Kr1]. Our proof based on boundaryvalues of analytic functions is from [Lin1] and [Sim1]. Another proof using Jensen’sinequality is given in [Be, Theorem 4.1]. A generalization of Krein’s conditionand a discrete analogue were obtained by H.L. Pedersen [Pd2]. The denseness ofpolynomials in Lp.R; �/ and Carleman’s condition are studied in [BC1], [BC2],[Ks], [KMP], [Bk1]. [BR], [If]. An index of determinacy for determinate measuresis defined in [BD]. Further elaborations of the determinacy problem can be found in[Lin2], [Stv], [SKv].

An interesting characterization of determinacy was discovered by C. Berg, Y.Chen, and M.E.H. Ismail [BCI], see also [BS1] for more refined results: Let �ndenote the smallest eigenvalue of the Hankel matrix Hn.s/ of a moment sequence s.Then s is determinate if and only if limn!1 �n D 0.

Chapter 5Orthogonal Polynomials and Jacobi Operators

In the preceding chapters we derived basic existence and uniqueness results formoment problems. In this chapter we develop two powerful tools for a “finer”study of one-dimensional moment problems: orthogonal polynomials and Jacobioperators.

Throughout this chapter we assume that s D .sn/n2N0 is a fixed positivedefinite moment sequence. The positive definiteness is crucial for the constructionof orthogonal polynomials and subsequent considerations. By Proposition 3.11 amoment sequence s is positive definite if and only if all Hankel determinants

Dn � Dn.s/

ˇˇˇˇˇˇ

s0 s1 s2 : : : sns1 s2 s3 : : : snC1s2 s3 s4 : : : snC2: : : : : : : : : : : : : : :

sn snC1 snC2 : : : s2n

ˇˇˇˇˇˇ; n 2 N0; (5.1)

are positive. We shall retain the notation (5.1) in what follows.In Sect. 5.1 we define and study orthogonal polynomials associated with s.

There are two distinguished sequences of orthogonal polynomials, the sequence. pn/n2N0 of orthonormal polynomials with positive leading coefficients (5.3) andthe sequence .Pn/n2N0 of monic orthogonal polynomials (5.6). In Sect. 5.2 wecharacterize these sequences in terms of three term reccurence relations and deriveFavard’s theorem (Theorem 5.10). The three term reccurence relation for . pn/n2N0

implies that the multiplication operator X is unitarily equivalent to a Jacobi operator(Theorem 5.14). The interplay of moment problems and Jacobi operators is studiedin Sect. 5.3.

Orthogonal polynomials of the second kind are investigated in Sect. 5.4. InSect. 5.5 the Wronskian is defined and some useful identities on the orthogonalpolynomials are derived. Section 5.6 contains basic results about zeros of orthogonalpolynomials, while Sect. 5.7 deals with symmetric moment problems.


93

94 5 Orthogonal Polynomials and Jacobi Operators

The study of orthogonal polynomials is an important subject that is of interest initself. We therefore develop some of the beautiful classical results and formulas inthis chapter, even if not all of them are used for the moment problem.

5.1 Definitions of Orthogonal Polynomialsand Explicit Formulas

Since the sequence s is positive definite, the equation

h p; qis WD Ls. p q/; p; q 2 CŒx�; (5.2)

defines a scalar product h�; �is on the vector space CŒx�. (Indeed, it is obvious thath�; �is is a positive semidefinite sesquilinear form. If h p; pis DPk;l skClckcl D 0 forsome p.x/ D P

k ckxk 2 CŒx�, then ck D 0 for all k and hence p D 0, since s is

positive definite. This proves that h�; �is is a scalar product.)Note that h p; qis is real for p; q 2 RŒx�, because the sequence s D .sn/ is real.The following orthonormal basis of the unitary space .CŒx�; h�; �is/ will play a

crucial role in what follows.

Proposition 5.1 There exists an orthonormal basis . pn/n2N0 of the unitary space.CŒx�; h�; �is/ such that each polynomial pn has degree n and a positive leading coef-ficient. The basis . pn/n2N0 is uniquely determined by these properties. Moreover,pn 2 RŒx�.

Proof For the existence it suffices to apply the Gram–Schmidt procedure to the basisf1; x; x2; : : : g of the unitary space .CŒx�; h�; �is/. Since the scalar product is real onRŒx�, we obtain an orthonormal sequence pn 2 RŒx� such that deg . pn/ D n. Uponmultiplying by �1 if necessary the leading coefficient of pn becomes positive. Theuniqueness assertion follows by a simple induction argument. ut

That the sequence . pn/n2N0 is orthonormal means that

h pk; pnis D ık;n for k; n 2 N0:

Definition 5.2 The polynomials pn; n 2 N0; are called orthogonal polynomialsof the first kind, or orthonormal polynomials, associated with the positive definitesequence s.

The existence assertion from Proposition 5.1 will be reproved by Proposition 5.3.We have included Proposition 5.1 in order to show that proofs are often much shorterif no explicit formulas involving the moments are required. This is true for manyother results as well, such as the recurrence relations (5.9) and (5.11). But our aimin this book is to provide explicit formulas for most quantities if possible.

5.1 Definitions of Orthogonal Polynomials and Explicit Formulas 95

Proposition 5.3 Set D�1 D 1. Then p0.x/ D 1ps0and for n 2 N and k 2 N0,

pn.x/ D 1pDn�1Dn

ˇˇˇˇˇˇˇˇ

s0 s1 s2 : : : sn

s1 s2 s3 : : : snC1s2 s3 s4 : : : snC2: : :

sn�1 sn snC1 : : : s2n�11 x x2 : : : xn

ˇˇˇˇˇˇˇˇ

; (5.3)

hxn; pnis DpDn=Dn�1 and hxk; pnis D 0 if k < n: (5.4)

The leading coefficient of pn ispDn�1=Dn . In particular, p1.x/ D s0x�s1p

s0.s0s2�s21/.

Proof Obviously, p0.x/ D 1ps0: In this proof let pn denote the polynomial (5.3).

First we verify (5.4). The polynomial xkpn.x/ is obtained by multiplying the lastrow of the determinant in (5.3) by xk. Applying the functional Ls to xkpn means thateach terms xkCj in the last row has to be replaced by skCj. Thus,

hxk; pnis D Ls.xkpn/ D 1p

Dn�1Dn

ˇˇˇˇˇˇˇˇ

s0 s1 s2 : : : sn

s1 s2 s3 : : : snC1s2 s3 s4 : : : snC2: : :

sn�1 sn snC1 : : : s2n�1sk skC1 skC2 : : : skCn

ˇˇˇˇˇˇˇˇ

: (5.5)

If k < n, the last row coincides with the .kC1/-th row in (5.5), so that hxk; pnis D 0.If k D n, the determinant in (5.5) is just the Hankel determinant Dn, that is,

hxn; pnis D 1pDn�1Dn

Dn DpDn=Dn�1;

which completes the proof of (5.4).Next we prove that h pk; pnis D ıkn. First let k < n. Since deg pk D k < n, we

conclude from (5.4) that h pk; pnis D 0. Similarly, h pk; pnis D 0 for k > n. Now letk D n. Since hxj; pnis D 0 for j < n, h pn; pnis is equal to hxn; pnis multiplied bythe leading coefficient of pn. From (5.3) it follows that pn has the leading coefficient

1pDn�1Dn

Dn�1 DpDn�1=Dn. Since hxn; pnis D

pDn=Dn�1 by (5.4), this yields

h pn; pnisD1.


From the uniqueness assertion of Proposition 5.1 it follows that the polynomialpn defined by (5.3) concides with the polynomial pn in Proposition 5.1. ut

We now define the general notion of orthogonal polynomials.

Definition 5.4 A sequence .Rn/n2N0 is called a sequence of orthogonal polynomi-als, briefly an OPS, with respect to s if Rn 2 RŒx�, degRn D n, and

hRk;Rnis D 0 for k ¤ n; k; n 2 N0:

Let .Rn/n2N0 be an OPS. Then kRnks ¤ 0, since h�; �is is a scalar product, and�nRn has positive leading term for �n D C or �n D �. Hence, by the uniqueness ofthe orthonormal sequence . pn/, we have �nkRnk�1n Rn D pn for all n 2 N0.

While there are many OPS for a given s, there is a unique OPS consisting ofmonic polynomials. Such an OPS will be called monic. Recall that a polynomialP of degree n is monic if its coefficient of xn is 1. Since pn has the leading termpDn�1=Dn; the polynomial

Pn.x/ WDpDn=Dn�1 pn.x/ D 1

Dn�1

ˇˇˇˇˇˇˇˇ

s0 s1 s2 : : : sn

s1 s2 s3 : : : snC1s2 s3 s4 : : : snC2: : :

sn�1 sn snC1 : : : s2n�11 x x2 : : : xn

ˇˇˇˇˇˇˇˇ

; n 2 N; (5.6)

is monic. Set P0.x/ D 1. Then .Pn/n2N0 is the unique monic OPS for s. Thus, thereare two distinguished sequences of orthogonal polynomials associated with s, theorthonormal sequence . pn/n2N0 and the monic OPS .Pn/n2N0 given by the aboveformulas (5.3) and (5.6), respectively.

We close this section with a beautiful classical formula for the polynomial Pn. Itwill be not used later in this book. Another formula for Pn is given in Lemma 6.27(i).

Proposition 5.5 (Heine Formulas) Suppose that � 2Ms. Then, for n 2 N, n � 2,we have

Pn.x/ D 1

nŠ Dn�1

ZRn

nYjD1.x � xj/

Y1�k<l�n

.xk � xl/2 d�.x1/ : : : d�.xn/; (5.7)

Dn�1 D 1

nŠ

ZRn

Y1�k<l�n

.xk � xl/2 d�.x1/ : : : d�.xn/: (5.8)

Proof Let us abbreviate QPn.x/ WD Dn�1pn.x/.

5.1 Definitions of Orthogonal Polynomials and Explicit Formulas 97

If � is a permutation of f0; 1; : : : ; n � 1g, we compute

QPn.x/ DZRn

d�.x�.0// : : : d�.x�.n�1//

ˇˇˇˇˇˇ

1 x1�.0/ : : : xn�.0/

x1�.1/ x2�.1/ : : : xnC1�.1/

: : : : : : : : : : : :

xn�1�.n�1/ x

n�.n�1/ : : : x2n�1

�.n�1/1 x : : : xn

ˇˇˇˇˇˇ

DZ

d�.x�.0// : : : d�.x�.n�1//

ˇˇˇˇˇˇ

1 x1�.0/ : : : xn�.0/1 x1�.1/ : : : xn�.1/: : : : : : : : : : : :

1 x1�.n�1/ : : : xn�.n�1/1 x : : : xn

ˇˇˇˇˇˇx0�.0/x

1�.1/ : : : x

n�1�.n�1/

DZ

d�.x0/ : : : d�.xn�1/

ˇˇˇˇˇˇ

1 x10 : : : xn01 x11 : : : xn1: : : : : : : : : : : :

1 x1n�1 : : : xnn�11 x : : : xn

ˇˇˇˇˇˇ.sign �/x0�.0/x

1�.1/ : : : x

n�1�.n�1/:

Here the first equality follows from formula (5.6) by replacing the moments sl in rowjC1 by

RRxl�. j/ d�.x

l�. j// and using the multilinearity of the determinant. For the

third equality the rows in the determinant are permuted and the integration variablesare changed to x0; : : : ; xn�1.

Summing over all nŠ permutations � of f0; 1; : : : ; n � 1g and inserting thedeterminant definition it follows that the polynomial nŠ QPn.x/ is equal to

ZRn

d�.x0/ : : : d�.xn�1/

ˇˇˇˇˇˇ

1 x10 : : : xn01 x11 : : : xn1: : : : : : : : : : : :

1 x1n�1 : : : xnn�11 x : : : xn

ˇˇˇˇˇˇ�

ˇˇˇˇ

1 x10 : : : xn�10

1 x11 : : : xn�11

: : : : : : : : : : : :

1 x1n�1 : : : xnn�1

ˇˇˇˇ :

Both determinants under the integral are Vandermonde determinants. The formula

ˇˇˇˇ

1 y10 : : : ym01 y11 : : : ym1: : : : : : : : : : : :

1 y1m : : : ymm

ˇˇˇˇ D

Y0�k<l�m

.yk � yl/


for a Vandermonde determinant implies that these determinants are

n�1YjD0.x � xj/

Y0�k<l�n�1

.xk � xl/ andY

0�k<l�n�1.xk � xl/;

respectively. Changing the variables from x0; : : : ; xn�1 to x1; : : : ; xn and inserting thepreceding expressions shows that nŠ QPn.x/ D nŠDn�1pn.x/ is equal to the integral in(5.7). This proves (5.7). Comparing the coefficients of xn in (5.7) gives (5.8). ut

5.2 Three Term Recurrence Relations

Orthogonal polynomials can be characterized and studied by means of three termrecurrence relations. We begin with the orthonormal sequence . pn/n2N0 .

Proposition 5.6 Set an D pDn�1DnC1D�1n and bn D Ls.xp2n/ for n 2 N0. Then we

have an > 0 and bn 2 R for n 2 N0, and

xpn.x/ D anpnC1.x/C bnpn.x/C an�1pn�1.x/; n 2 N0; (5.9)

where a�1 WD 1 and p�1.x/ WD 0. In particular, p0.x/ D s�1=20 ,

p1.x/ D s�1=20 a�1

0 .x�b0/; p2.x/ D s�1=20 .a�1

0 a�11 .x�b0/.x�b1/� a0a

�11 /:

Proof Since Dk > 0 and pk 2 RŒx� for all k, we have an > 0 and bn 2 R.Since xpn.x/ has degree nC1 and f p0; : : : ; pnC1g is a basis of the space of real

polynomials of degree less than or equal to nC1, there are reals cnk such thatxpn.x/ D PnC1

kD0 cnkpk.x/. Comparing the coefficients of xnC1, it follows that cn;nC1is the quotient of the leading coefficients of pn and pnC1. By Proposition 5.3 thisyields cn;nC1 D an. Because the basis f pkg is orthonormal, we have

cnk D hxpn; pkis D Ls.xpnpk/ D h pn; xpkis; k D 0; : : : ; nC 1: (5.10)

Since xpk is in the span of p0; : : : ; pkC1, (5.10) implies that cnk D 0 when kC1 < n.Further, cn;n D Ls.xp2n/ D bn by (5.10). Using that cn�1;n is real we derive

cn;n�1 D h pn; xpn�1is D˝pn;

nXkD0

cn�1;kpk˛sD cn�1;n:

Hence an�1 D cn;n�1. Putting the preceding together we have proved (5.9).The formulas for p1 and p2 are easily computed from (5.9). ut

Corollary 5.7 The leading term of pn.x/ isq

Dn�1

DnD s�1=2

0

Qn�1kD0 a�1

k for n 2 N.

5.2 Three Term Recurrence Relations 99

Proof We proceed by induction using the relation ak D pDk�1DkC1 D�1k , k 2 N0.

For n D 1 we haveq

D0D1D D�1=2

0 .pD�1D1D�1

0 /�1 D s�1=2

0 a�10 .

If the assertion holds for n 2 N, then

sDn

DnC1D Dnp

Dn�1DnC1

sDn�1DnD a�1

n s�1=20

n�1YkD0

a�1k D s�1=2

0

nYkD0

a�1k : ut

Equation (5.9) is a three term recurrence relation for the polynomials pn, that is,the polynomial pnC1 is determined by

pnC1.x/ D .x � bn/a�1n pn.x/� an�1a�1

n pn�1.x/; n 2 N0; (5.11)

where a�1 WD 1 and p�1.x/ WD 0. Hence, if pj�1.x/ and pj.x/ are given, thenEq. (5.9) (and likewise (5.11)) determines all polynomials pn.x/; n � jC1; uniquely.This property of a three term reccurance relation will often be used.

The formula an D pDn�1DnC1D�1n expresses an in terms of determinants

involving only the moments. To derive a similar result for the numbers bn we set

�n D

ˇˇˇˇˇˇ

s0 s1 s2 : : : sn�1 snC1s1 s2 s3 : : : sn snC2s2 s3 s4 : : : snC1 snC3: : :

sn snC1 snC2 : : : s2n�1 s2nC1

ˇˇˇˇˇˇ; n 2 N; and �0 D s1; (5.12)

that is, �n is obtained from the Hankel determinant Dn (see (5.1)) by adding 1 to allindices in the last column.

Proposition 5.8 For any n;m 2 N0, we have

bnC1 D �nC1DnC1

� �n

Dnand

mXnD0

bn D �m

Dm: (5.13)

Proof First let m D 0. Then .x � b0/p0 D a0p1 by (5.6). Since Ls. p1/ D 0, we getp0s1 D Ls. p0x/ D Ls. p0b0/ D p0b0s0 and hence b0 D s1s�1

0 D �0D�10 .

Let n 2 N. By developing the determinant in (5.3) after the last row it followsthat the coefficients of xn and xn�1 are Dn and��n�1, respectively, so the coefficient

of xn and xn�1 in pn are 1pDn�1Dn

Dn Dq

Dn�1

Dn; and � 1p

Dn�1Dn�n�1, respectively.

Recall that an D pDn�1DnC1D�1n by Proposition 5.6. Comparing the coeffients of

xn on both sides of the Eq. (5.9) and inserting an D pDn�1DnC1D�1n we obtain

� 1pDn�1Dn

�n�1 D �pDn�1DnC1

Dn

1pDnDnC1

�n C bn

sDn�1Dn

:


From this equation it follows that

bn D �n

Dn� �n�1

Dn�1for n 2 N;

which proves the first equality of (5.13). Therefore, since b0 D �0D�10 , we derive

mXnD0

bn D b0 CmX

nD1

��n

Dn� �n�1

Dn�1

�D b0 C �m

Dm� �0

D0D �m

Dm: ut

The next proposition contains the three term recurrence relation for the monicOPS. Using the formula Pn D

pDn=Dn�1 pn (by (5.6)) it is easily derived from the

recurrence relation (5.9) for pn. We omit the details of these simple computations.

Proposition 5.9 Let .Pn/n2N0 be the monic OPS for s, see (5.6). Let an�1 and bn beas in Proposition 5.6 and P�1 WD 0. Then

PnC1.x/ D .x � bn/Pn.x/� a2n�1Pn�1.x/; n 2 N0: (5.14)

In particular, P0.x/ D 1, P1.x/ D x � b0, and P2.x/ D .x � b0/.x � b1/� a20:

The next result is Favard’s theorem. It is a converse to Proposition 5.9. Its maindirection states that for each set of parameters fan; bn W n 2 N0g with an > 0 andbn 2 R the recurrence relation (5.14) defines a monic OPS of some positive definitesequence s and hence of some measure � 2Ms:

Theorem 5.10 Let .˛n/n2N0 and .ˇn/n2N0 be complex sequences and set ˛�1 WD1. Let .Rn/n2N0 denote the sequence of monic polynomials Rn which is uniquelydetermined by the relations

RnC1.x/ D .x � ˇn/Rn.x/� ˛n�1Rn�1.x/; n 2 N0; (5.15)

R�1.x/ D 0 and R0.x/ D 1: (5.16)

There exists a positive definite real sequence s such that .Rn/n2N0 is the monic OPSfor s if and only if ˛n > 0 and ˇn 2 R for all n 2 N0. If s0 is a given positivenumber, then this sequence s D .sn/n2N0 is uniquely determined.

Further, if ˛n > 0 and ˇn 2 R for n 2 N0 and s0 > 0 are given, then there existsa measure � 2Ms such that �.R/ D s0 and for j; k 2 N0, j ¤ k, and n 2 N,

ZR

Rj.x/Rk.x/ d�.x/ D 0 ;Z

R2n.x/d�.x/ D ˛n�1˛n�2 : : : ˛0s0: (5.17)

Proof First note that the sequence .Rn/ is indeed uniquely determined by (5.15) and(5.16). Clearly, Rn is monic and has degree n. Hence .Rn/n2N0 is a vector space basisof CŒx�, so we can define a linear functional L on

L.R0/ D L.1/ D s0 and L.Rn/ D 0 for n 2 N: (5.18)

5.2 Three Term Recurrence Relations 101

Applying L to both sides of (5.15) we get L.xRn/ D 0 for n � 2. Multiplying (5.15)by x, applying L again by using the latter equality, we obtain L.x2Rn/ D for n � 3.Proceeding in this manner we derive

L.xjRn/ D 0 for j D 0; : : : ; n � 1; n 2 N: (5.19)

Since degRm D m, (5.19) implies that

L.RjRk/ D 0 for j; k 2 N0; j ¤ k: (5.20)

Multiplying (5.15) by xn�1 and applying L by using (5.19) again we get L.xnRn/ D˛n�1L.xn�1Rn�1/ and hence L.xnRn/ D ˛n�1˛n�2 : : : ˛0s0, because L.R0/ D s0.Since Rn is monic, the preceding combined with (5.19) yields

L.R2n/ D ˛n�1˛n�2 : : : ˛0s0; n 2 N: (5.21)

After all these technical preparations we are able to prove the assertions.First suppose that .Rn/ is an OPS for some positive definite real sequence s.

Then Ls is equal to the functional L defined by (5.18) with s0 D Ls.1/ > 0, sincehRn;R0is D Ls.Rn/ D 0 for n 2 N. By the definition of an OPS, Rn 2 RŒx�, so thathRn;Rnis D Ls.R2n/ > 0 for n 2 N. Hence, by (5.21), ˛n > 0 for all n 2 N. Since˛n > 0 and Rn 2 RŒx�, it follows from (5.15) that ˇn is real.

Conversely, assume that ˛n > 0 and ˇn 2 R for n 2 N0. Let s0 > 0 be arbitrary.Then Rn 2 RŒx� and L.R2n/ > 0 by (5.21). Combined with (5.20) the latter impliesthat the linear functional L defined by (5.18) is a positive functional on CŒx� suchthat L. pp/ > 0 for all p 2 CŒx�, p ¤ 0. Therefore, setting sn WD L.xn/, n 2 N0, weget a positive definite sequence s D .sn/n2N0 . By (5.20), .Rn/ is an OPS for s. Since.Rn/ is monic, it is the unique monic OPS associated with s.

Further, by Theorem 3.8, there exists a measure � 2 MC.R/ such that sn DRxnd� for n 2 N0. Then �.R/ D L.1/ D s0: Since L D Ls and � is a representing

measure for s, the equations (5.17) follow from (5.20) and (5.21). utRemark 5.11

1. Clearly, if .Rn/ is the monic OPS for a positive definite sequence s, so it is foreach positive multiple of s. Hence the number s0 D Ls.1/ cannot be determinedfrom the monic OPS .Rn/.

2. Favard’s theorem for the recurrence relation (5.7) is contained in Theorem 5.14below, see e.g. Remark 5.15 in the next section. ıComparing (5.14) and (5.15) shows that ˛n D a2n for the monic OPS .Pn/, so

(5.21) yields the following formula. It also follows from Corollary 5.7.

Corollary 5.12 kPnk2s D Ls.P2n/ D a2n�1a2n�2 : : : a20s0 for n 2 N.


5.3 The Moment Problem and Jacobi Operators

The three term recurrence relation (5.9) links the moment problem to Jacobioperators. These operators are the basic objects for the operator-theoretic approachto the moment problem. This approach will be elaborated in the subsequent chapters.

Let Hs denote the Hilbert space completion of the unitary space .CŒx�; h�; �is/ andX the multiplication operator by the variable x with domain CŒx� on Hs :

Xp.x/ WD xp.x/ for p 2 D.X/ WD CŒx�:

Then X is a densely defined symmetric operator with domain CŒx� on the Hilbertspace Hs, since

hXp; qis D Ls.xp q/ D Ls. p xq/ D h p;Xqis for p; q 2 CŒx�:

Let fen W n 2 N0g be the standard orthonormal basis of the Hilbert space l2.N0/

given by enWD.ıkn/k2N0 . Since f pn W n 2 N0g is an orthonormal basis of Hs, there isa unitary isomorphism U of Hs onto l2.N0/ defined by Upn D en. Then, by (5.9),T WD UXU�1 is a symmetric operator on l2.N0/ which acts by

Ten D anenC1 C bnen C an�1en�1; n 2 N0; (5.22)

where e�1 WD 0. The domain D.T/ D U.CŒx�/ is the linear span of vectors en, thatis, D.T/ is the vector space d of finite complex sequences .�0; : : : ; �n; 0; : : : /. Forany finite sequence � D .�n/ 2 d we obtain

T�X

n�nen

� DXn�n.anenC1 C bnen C an�1en�1/

DX

n.�n�1an�1 C �nbn C �nC1an/en

DX

n.an�nC1 C bn�n C an�1�n�1/en ;

where we have set ��1 WD 0. That is,

.T�/0 D a0�1 C b0�0; .T�/n D an�nC1 C bn�n C an�1�n�1; n 2 N: (5.23)

Equation (5.23) means that the operatorT acts on a sequence � 2 d by multiplicationwith the infinite matrix

J D

0BBBBB@

b0 a0 0 0 0 : : :

a0 b1 a1 0 0 : : :

0 a1 b2 a2 0 : : :

0 0 a2 b3 a3 : : :

: : : : : :: : :

: : :: : :

1CCCCCA: (5.24)

5.3 The Moment Problem and Jacobi Operators 103

A matrix J of this form is called a (semi-finite) Jacobi matrix. It is symmetric andtridiagonal. The corresponding operator T � TJ (likewise its closure) is called aJacobi operator. Thus we have shown that the multiplication operator X is unitarilyequivalent to the Jacobi operator TJ for the matrix (5.24).

The numbers sns�10 can be recovered from the Jacobi operator T by

sns�10 D s�1

0 hxn1; 1is D h.X/np0; p0is D hTne0; e0i; n 2 N0: (5.25)

That is, sns�10 is the entry in the left upper corner of the matrix Jn. Thus, if s0 D 1,

then all moments sn are uniquely determined from the Jacobi matrix. In particular,

s1s�10 D hTe0; e0i D b0; s2s

�10 D hT2e0; e0i D b20 C a20;

s3s�10 D hT3e0; e0i D b30 C 2a20b0 C a20b1:

Remark 5.13 Let Qs D cs, where c > 0, be a multiple of the positive definitesequence s. Then s and Qs have the same Jacobi matrix J (see Exercise 5.1) andthe same Jacobi operator T. Also, the multiplication operators X for Qs and s areunitarily equivalent and we have � 2 Ms if and only if c� 2 MQs. Thus for allself-adjointness problems of X and T and for the determinacy of s we could assumethat s0 D 1. ı

Conversely, let a Jacobi matrix (5.24) be given, where an; bn 2 R and an > 0.We define a linear operator T with domain D.T/ D d on the Hilbert space l2.N0/

by (5.23). Clearly, T is a symmetric operator and TD.T/ D.T/. Set

s D .snWDhTne0; e0i/n2N0 : (5.26)

We prove that the sequence s is a positive definite. For �0; : : : ; �n 2 C and n 2 N0,

nXk;lD0

skCl�k�l DnX

k;lD0hTkCle0; e0i�k�l D

��nX

lD0�lT

le0

��2

� 0: (5.27)

Hence s is positive semidefinite. By a simple induction argument we show thatTle0 � a0 : : : al�1el 2 Lin fe0; : : : ; el�1g for l 2 N. Therefore, if the expressionin (5.27) is zero, we conclude that all �n are zero, since aj ¤ 0 for all j. Thus, s is apositive definite real sequence. By (5.26) we have s0 D hT0e0; e0is D 1.

Recall that U is the unitary of Hs onto l2.N0/ defined by Upn D en; n 2 N0.Put Qpn WD U�1en for n 2 N0. Clearly, (5.23) implies (5.22), so the polynomialsQpn satisfy the recurrence relation (5.9). Since an > 0 and bn 2 R by assumption,it follows from (5.9) that Qpn 2 RŒx� has degree n and positive leading term. Thepolynomials Qpn are orthonormal in Hs, since the vectors en are in l2.N0/. Therefore,the Qpn are the polynomials of the first kind for s, that is, pn D Qpn for n 2 N0. HenceT is the Jacobi operator associated with s according to the preceding construction.


Summarizing, we have proved the following

Theorem 5.14 The preceding construction provides a one-to-one correspondencebetween positive definite sequences s satisfying s0 D 1 and Jacobi matrices J of theform (5.24), where bn 2 R and an > 0 for n 2 N0, and a unitary equivalencebetween the multiplication operator X and the Jacobi operator T � TJ given bya unitary U such that Upn D en; n 2 N0. The Jacobi parameters an and bn in thematrix (5.24) are the numbers occuring in the three term recurrence relation (5.9).

For notational simplicity we identify the operators X and T D UXU�1 in whatfollows, where U is the unitary of Hs onto l2.N0/ defined by Upn D en, n 2 N0.

Thus, by Theorem 5.14, for any positive definite sequence s the multiplicationoperator X is unitarily equivalent to the Jacobi operator TJ of the matrix (5.24),where an > 0 and bn 2 R are as in Proposition 5.6. Conversely, if J is a Jacobimatrix (5.24) with an > 0 and bn 2 R, then (5.26) is a positive definite sequence forwhich the above procedure leads again to the matrix (5.24).

Remark 5.15 Theorem 5.14 can be considered as Favard’s theorem. Indeed, sup-pose that numbers bn 2 R and an > 0, n 2 N0; are given. Then, by Theorem 5.14,there is a positive definite sequence s with s0 D 1 such that the Jacobi operatorT D TJ is unitarily equivalent to the multiplication operator X in .CŒx�; h�; �is/ andthe polynomials pn defined by (5.7) form the corresponding orthonormal OPS for s.Further, by Hamburger’s Theorem 3.8, there is a Radon measure � 2MC.R/ suchthat sn � hTne0; e0i D

Rxn d� for n 2 N0. ı

5.4 Polynomials of the Second Kind

In this section, we assume that an and bn, n 2 N0, denote the Jacobi parameters ofthe positive definite sequence s and we set a�1 WD 1. Our aim is to develop anothersequence .qn/n2N0 of polynomials associated with s.

For a complex sequence � D .�n/n2N0 we define a complex sequence T � by

.T �/n D an�nC1 C bn�n C an�1�n�1 for n 2 N0; where ��1 WD 0: (5.28)

Then T is a linear mapping of the vector space of all complex sequences. For � 2 d,it is obvious that T � 2 d and T � D T� , where T D TJ is the Jacobi operator.

Suppose that z 2 C and consider the three term recurrence relation

.T �/n � an�nC1 C bn�n C an�1�n�1 D z�n; (5.29)

where ��1 WD 0, for an arbitrary complex sequence � D .�n/n2N0 . Clearly, sincean > 0, if we fix two initial data �k�1 and �k and assume that the relation (5.29) issatisfied for all n � k, all terms �n, where n � kC 1, are uniquely determined.

5.4 Polynomials of the Second Kind 105

We set ��1 D 0, �0 D s�1=20 and assume that (5.29) holds for all n 2 N0.

Comparing (5.29) with (5.9) by using that p�1.z/ D 0 and p0.z/ D s�1=20 we

conclude that �n is the value pn.z/ of the polynomial pn from Proposition 5.1. Weabbreviate

pz WD . p0.z/; p1.z/; p2.z/; : : : /; z 2 C: (5.30)

Now we set �0 D 0, �1 D a�10 s1=20 and suppose that (5.29) holds for all n 2 N.

(Note that we do not assume (5.29) for n D 0.) The numbers �n are then uniquelydetermined and we denote �n by qn.z/, n 2 N0. Clearly, the same solution isobtained if we start with the initial data ��1 D �s1=20 , �0 D 0 and require (5.29)for all n 2 N0.

Using relation (5.29) it follows easily by induction on n that qn.z/, n 2 N, is apolynomial in z of degree n�1. We denote the corresponding sequence by

qz WD .q0.z/; q1.z/; q2.z/; : : : /; z 2 C: (5.31)

By definition, q0.z/ D 0 and q1.z/ D a�10 s1=20 . Further, q2.z/ D .z�b1/.a0a1/�1s1=20 :

It should be emphasized that the numbers an; bn and the polynomials pn; qndepend only on the sequence s, but not on any representing measure.

Lemma 5.16 T pz D zpz and T qz D s1=20 e0 C zqz for all z 2 C.

Proof By the recurrence relations we have .T pz/n D zpn.z/ � .zpz/n for n 2 N0

and .T qz/n D zqn.z/ � .zqz/n for n 2 N. Using that �0D0 and �1Da�10 s1=20 we

compute the zero component .T qz/0 by

.T qz/0 D a0�1 C b0�0 D a0a�10 s1=20 C 0 D s1=20 C z�0 D s1=20 C zq0.z/: ut

Definition 5.17 The polynomials qn.z/; n 2 N0; are the orthogonal polynomialsof the second kind associated with the positive definite sequence s.

From the defining relations and intial data for qn.z/ it follows that the polynomi-alsepn.z/ WD s�1=2

0 a0qnC1.z/ satisfy the recurrence relation (5.29) with an replacedbyean WD anC1 and bn byebn WD bnC1 and the initial dataep�1 D 0,ep0 D 1. Therefore,by Theorem 5.14, the polynomialsepn.z/, n 2 N0, are the orthonormal polynomialsof first kind with respect to the shifted Jacobi matrix

eJ D

0BBBBB@

b1 a1 0 0 0 : : :

a1 b2 a2 0 0 : : :

0 a2 b3 a3 0 : : :

0 0 a3 b4 a4 : : :

: : : : : :: : :

: : :: : :

1CCCCCA: (5.32)


The corresponding positive definite sequence ises D .esn/, whereesnDheTne0; e0i andeT is the Jacobi operator corresponding to the Jacobi matrixeJ. Hence qn; n 2 N; areorthogonal polynomials (according to Definition 5.4) for the sequence Qs.

The next proposition contains another useful description of the polynomials qn.Let r.x/ DPn

kD0 �kxk 2 CŒx�, n 2 N, be a polynomial. For any fixed z 2 C,

r.x/� r.z/

x � zD

nXkD0

�kxk � zk

x � zD

nXkD1

k�1XlD0

�kzk�lxl

is also a polynomial in x, so we can apply the functional Ls to it. We shall write Ls;xto indicate that x is the corresponding variable.

Proposition 5.18 qn.z/ D Ls;x�pn.x/�pn.z/

x�z

for n 2 N0 and z 2 C.

Proof Let us denote the polynomial on the right-hand side by rn.z/. From therecurrence relation (5.9) we obtain for n 2 N,

anpnC1.x/ � pnC1.z/

x � zC bn

pn.x/� pn.z/

x � zC an�1

pn�1.x/ � pn�1.z/x � z

D xpn.x/� zpn.z/

x � zD z

pn.x/ � pn.z/

x � zC pn.x/:

Applying the functional Ls;x to this identity and using the orthogonality relation0 D h pn; 1is D Ls;x. pn/ for n � 1 we get

anrnC1.z/C bnrn.z/C an�1rn�1.z/ D zrn.z/ for n 2 N:

Since p1.x/ D s�1=20 a�1

0 .x�b0/ and p0 D s�1=20 by Proposition 5.6, we have r0.z/ D

Ls;x.0/ D 0 D q0.z/ and

r1.z/ D Ls;x

�p1.x/ � p1.z/

x � z

�D s�1=2

0 a�10 Ls;x

�x � z

x � z

�D s�1=2

0 a�10 s0 D q1.z/:

This shows that the sequence .rn.z// satisfies the same recurrence relation and initialdata as .qn.z//. Therefore, rn.z/ D qn.z/ for n 2 N0. ut

The next corollary expresses the polynomials qn in terms of the moments.

Corollary 5.19 Set qn;k.z/ DPk�1lD0 sk�l�1zl for k � 1 and qn;0 D 0. For n 2 N,

qn.z/ WD 1pDn�1Dn

ˇˇˇˇˇˇˇ

s0 s1 s2 : : : sns1 s2 s3 : : : snC1s2 s3 s4 : : : snC2: : :

sn�1 sn snC1 : : : s2n�1qn;0.z/ qn;1.z/ qn;2.z/ : : : qn;n.z/

ˇˇˇˇˇˇˇ

; z 2 C: (5.33)

5.4 Polynomials of the Second Kind 107

Proof From (5.3) it follows that pn.x/�pn.z/x�z is given by the same expression as in

(5.3) when the entries xk of the last row are replaced by xk�zk

x�z DPk�1

lD0 xk�l�1zl:Applying the functional Ls;x to this determinant gives the determinant in (5.33). ByProposition 5.18, the corresponding polynomial in z is qn.z/. utCorollary 5.20 The monic polynomial associated with qn.z/ is

Qn.z/ WD s�10 Ls;x

�Pn.x/ � Pn.z/

x � z

�for n 2 N0: (5.34)

Further, for n 2 N and z 2 C we have

Pn.z/ D s1=20 a0a1 : : : an�1pn.z/; Qn.z/ D s�1=20 a0a1 : : : an�1qn.z/; (5.35)

qn.z/

pn.z/D s0

Qn.z/

Pn.z/: (5.36)

Proof Let cnxn be the leading term of pn.x/. Then Pn D c�1n pn. Therefore, by

(5.34) and Proposition 5.18, Qn.z/ D s�10 c�1

n qn.z/. Since xn�zn

x�z DPn�1

lD0 xn�l�1zl;the leading term of qn is zn�1cnLs;x.1/ D cns0zn�1. Hence Qn.z/ D s�1

0 c�1n qn.z/ is

monic.By Corollary 5.7, c�1

n D s1=20 a0a1 : : : an�1. Since Pn D c�1n pn and Qn D

s�10 c�1

n qn, this implies (5.35) and hence also (5.36). utIn what follows we will often use the function fz and the Stieltjes transform I�

(see Appendix A.2) of a finite Radon measure � on R. They are defined by

fz.x/ WD 1

x � zand I�.z/ WD

ZR

1

x � zd�.x/; z 2 CnR: (5.37)

Proposition 5.21 Suppose that � 2Ms. For z 2 CnR and n 2 N0,

h fz; pniL2.R;�/ D qn.z/C I�.z/pn.z/; (5.38)

kqz C I�.z/pzk2 D1XnD0jqn.z/C I�.z/pn.z/j2 � Im I�.z/

Im z: (5.39)

In particular, qzC I�.z/pz 2 l2.N0/. Moreover, we have equality in the inequality of(5.39) if and only if the function fz is inHs.

Proof Clearly, the bounded function fz.x/ D 1x�z is in L2.R; �/. We compute

h fz; pniL2.R;�/ DZR

pn.x/�pn.z/x�z d�.x/C

ZR

pn.z/

x�z d�.x/

D Ls;x

�pn.x/�pn.z/

x�z�C pn.z/I�.z/ D qn.z/C I�.z/pn.z/;


which proves (5.38). Here the equality before last holds, since � is a representingmeasure for s, and the last equality follows from Proposition 5.18.

The equality in (5.39) is merely the definition of the norm of l2.N0/. Since.CŒx�; h�; �is/ is a subspace of L2.R; �/, f pn W n 2 N0g is an orthonormal subsetof L2.R; �/. The inequality in (5.39) is just Bessel’s inequality for the Fouriercoefficients of fz with respect to this orthonormal set, because

k fzk2L2.R;�/ DZ

1

jx�zj2 d�.x/ DZ

1

z�z�1

x�z �1

x�z�d�.x/ D Im I�.z/

Im z:

By an elementary fact from Hilbert space theory, equality in Bessel’s inequalityholds if and only if fz belongs to the closed subspace generated by the polynomialspn, that is, fz 2 Hs. utCorollary 5.22 pz 2 l2.N0/ if and only if qz 2 l2.N0/ for z 2 CnR.

Proof Let z 2 CnR. Since then I�.z/ ¤ 0 and qz C I�.z/pz 2 l2.N0/ byProposition 5.21, it is clear that pz 2 l2.N0/ is equivalent to qz 2 l2.N0/. ut

5.5 The Wronskian and Some Useful Identities

Suppose that � D .�n/n2N0 and ˇ D .ˇn/n2N0 are complex sequences. We definetheir Wronskian as the sequence W.�; ˇ/ D .W.�; ˇ/n/n2N0 with terms

W.�; ˇ/n WD an.�nC1ˇn � �nˇnC1/; n 2 N0; (5.40)

Let T be the linear mapping of complex sequences defined by (5.28), that is,

.T �/n D an�nC1 C bn�n C an�1�n�1 for n 2 N0; where ��1 WD 0:

The following lemma on the Wronskian is the crux for several applications. It willplay an essential role in the study of the adjoint of the Jacobi operator in Sect. 6.2.

Lemma 5.23 Let � D .�n/n2N0 , ˇ D .ˇn/n2N0 be sequences, and x; z 2 C. Then

nXkD0

�.T �/kˇk � �k.T ˇ/k/

� D W.�; ˇ/n for n 2 N0: (5.41)

Let m; n 2 N0; n > m: If .T �/k D x�k and .T ˇ/k D zˇk for k D mC1; : : : ; n, then

.x � z/nX

kDmC1�kˇk D W.�; ˇ/n �W.�; ˇ/m: (5.42)

5.5 The Wronskian and Some Useful Identities 109

In particular, if x D z in this case, then

W.�; ˇ/n D W.�; ˇ/m: (5.43)

Proof We prove the first identity (5.41) by computing

nXkD0

Œ.T �/kˇk � �k.T ˇ/k/�

D .a0�1 C b0�0/ˇ0 � �0.a0ˇ1 C b0ˇ0/

CnX

kD1Œ.ak�kC1Cbk�kCak�1�k�1/ˇk � �k.akˇkC1CbkˇkCak�1ˇk�1/�

D a0.�1ˇ0 � �0ˇ1/CnX

kD1Œ.ak.�kC1ˇk��kˇkC1/ � ak�1.�kˇk�1��k�1ˇk/�

DW.�; ˇ/0 CnX

kD1ŒW.�; ˇ/k �W.�; ˇ/k�1� D W.�; ˇ/n:

Equation (5.42) is obtained by applying (5.41) to n and m and taking thedifference of both sums. Setting x D z in (5.42) yields (5.43). ut

Now we use Lemma 5.23 to derive some important identities on the polynomialspn; qn. Further, we define four polynomials An;Bn;Cn;Dn that will be used later inSect. 7.1. Equation (5.47) is called the Christoffel–Darboux formula.

Proposition 5.24 For x; z 2 C and n 2 N0, we have

An.x; z/ WD .x�z/nX

kD0qk.x/qk.z/ D an.qnC1.x/qn.z/�qn.x/qnC1.z//; (5.44)

Bn.x; z/ WD �1C .x�z/nX

kD0pk.x/qk.z/ D an. pnC1.x/qn.z/�pn.x/qnC1.z//;

(5.45)

Cn.x; z/ WD 1C .x�z/nX

kD0qk.x/pk.z/ D an.qnC1.x/pn.z/�qn.x/pnC1.z//;

(5.46)

Dn.x; z/ WD .x�z/nX

kD0pk.x/pk.z/ D an. pnC1.x/pn.z/�pn.x/pnC1.z//: (5.47)

Proof All four identities are derived from Eq. (5.42). As samples we verify (5.45)and (5.47). Recall that the sequences px D . pn.x//, qz D .qn.z//, and qz D .qn.z//


satisfy the relations

.T px/n D xpn.x/; .T qz/n D zqn.z/; .T pz/n D zpn.z/ for n 2 N: (5.48)

Hence the assumptions of Eq. (5.42) in Lemma 5.23 with ˛ D px, ˇ D qz; pz, andm D 0 are satisfied. Inserting p0.x/ D s�1=2

0 , p1.x/ D s�1=20 a�1

0 .x�b0/, q0.z/ D 0,

and q1.z/ D a�10 s1=20 , we compute

W.px; qz/0 D a0. p1.x/q0.z/�p0.x/q1.z// D �1; (5.49)

W.px; pz/0 D a0. p1.x/p0.z/�p0.x/p1.z// D .x � z/s�10 : (5.50)

For n D 0 the right-hand sides of (5.45) and (5.47) are just W.px; qz/0 andW.px; pz/0, respectively, so these equations hold by (5.49) and (5.50). Now supposethat n 2 N. Then (5.42) applies with m D 0.

From equations (5.42) with ˛ D px; ˇ D qz by using (5.49) and (5.40) we derive

.x�z/nX

kD0pk.x/qk.z/ D .x�z/

nXkD1

pk.x/qk.z/

D W.px; qz/n�W.px; qz/0 D an. pnC1.x/qn.z/�pn.x/qnC1.z//C 1:

Applying (5.42) with ˛ D px; ˇ D pz combined with (5.50) and (5.40) yields

.x�z/nX

kD0pk.x/pk.z/ D .x�z/s�1

0 C .x�z/nX

kD1pk.x/pk.z/

D .x � z/s�10 CW.px; pz/n�W.px; pz/0 D an. pnC1.x/pn.z/�pn.x/pnC1.z//:

This proves (5.45) and (5.47). utCorollary 5.25 Let x 2 C and n 2 N0. Then

nXkD0

pk.x/2 D an

�p0nC1.x/pn.x/� p0

n.x/pnC1.x/�; (5.51)

1

anD pn.x/qnC1.x/ � pnC1.x/qn.x/: (5.52)

Proof From the identity (5.47) it follows that

nXkD0

pk.x/pk.z/ D anŒpnC1.x/ � pnC1.z/�pn.z/�Œpn.x/� pn.z/�pnC1.z/

x � z:

5.5 The Wronskian and Some Useful Identities 111

Letting z ! x we obtain (5.51). Equation (5.52) is obtained by setting x D z in(5.46). utCorollary 5.26 For x; z 2 C and n 2 N0,

pnC1.x/ � pnC1.z/ D .x � z/nX

kD0Œpk.z/qnC1.z/ � pnC1.z/qk.z/�pk.x/; (5.53)

qnC1.x/ � qnC1.z/ D .x � z/nX

kD0Œpk.z/qnC1.z/ � pnC1.z/qk.z/�qk.x/: (5.54)

Proof First we prove (5.53). We multiply (5.47) by qnC1.z/ and (5.45) by �pnC1.z/and add both equations. On the right we get

an�qnC1.z/ŒpnC1.x/pn.z/ � pn.x/pnC1.z/� � pnC1.z/ŒpnC1.x/qn.z/ � pn.x/qnC1.z/�

�DpnC1.x/ anŒqnC1.z/pn.z/� qn.z/pnC1.z/�

DpnC1.x/W.qz; pz/n D pnC1.z/W.qz; pz/0 D pnC1.x/: (5.55)

Here for the first equality the term pn.x/pnC1.z/qnC1.z/ cancels, while the secondis just the definition of W.qz; pz/n. By (5.48) the assumptions of Lemma 5.23 aresatisfied, so (5.43) yields W.qz; pz/n D W.qz; pz/0 for n 2 N. For n D 0 this istrivial. This gives the third equality. Since W.qz; pz/0 D 1, the last equality in (5.55)holds.

On the left we obtain

.x�z/nX

kD0qnC1.z/pk.x/pk.z/C pnC1.z/ � .x � z/

nXkD0

pnC1.z/pk.x/qk.z/

D pnC1.z/C .x � z/nX

kD0Œpk.z/qnC1.z/� pnC1.z/qk.z/�pk.x/: (5.56)

Comparing (5.55) and (5.56) we obtain (5.53).The proof of (5.54) is very similar by using (5.46) and (5.44) instead of (5.47)

and (5.45). First we multiply (5.46) by qnC1.z/ and (5.44) by�pnC1.z/. Adding bothequations, on the right-hand side we derive

an�qnC1.z/ŒqnC1.x/pn.z/ � qn.x/pnC1.z/� � pnC1.z/ŒqnC1.x/qn.z/ � qn.x/qnC1.z/�

�DqnC1.x/anŒqnC1.z/pn.z/� qn.z/pnC1.z/�

DqnC1.x/W.qz; pz/n D qnC1.x/W.qz; pz/0 D qnC1.x/:


The left-hand side gives

qnC1.z/C .x � z/nX

kD0qnC1.z/qk.x/pk.z/� .x � z/

nXkD0

pnC1.z/qk.x/qk.z/

D qnC1.z/C .x � z/nX

kD0Œpk.z/qnC1.z/� pnC1.z/qk.z/�qk.x/:

Now (5.54) follows by comparing the results on both sides. utCorollary 5.27 For any x; z 2 C and n 2 N0, we have

An.x; z/Dn.x; z/� Bn.x; z/Cn.x; z/ D 1; (5.57)

Dn.x; 0/Bn.z; 0/ � Bn.x; 0/Dn.z; 0/ D �Dn.x; z/: (5.58)

Proof Inserting the identities (5.44)–(5.47) from Proposition 5.24 we compute

An.x; z/Dn.x; z/ � Bn.x; z/Cn.x; z/

D anŒqnC1.x/qn.z/�qn.x/qnC1.z/� anŒ. pnC1.x/pn.z/�pn.x/pnC1.z/�

�anŒpnC1.x/qn.z/�pn.x/qnC1.z/� anŒqnC1.x/pn.z/�qn.x/pnC1.z/�

D a2n ŒpnC1.x/qn.x/� pn.x/qnC1.x/� ŒpnC1.z/qn.z/ � qnC1.z/pn.z/�

D Bn.x; x/Bn.z; z/ D .�1/.�1/ D 1:

Likewise, we derive

Dn.x; 0/Bn.z; 0/ � Bn.x; 0/Dn.z; 0/

D anŒpnC1.x/pn.0/�pn.x/pnC1.0/� anŒpnC1.z/qn.0/�pn.z/qnC1.0/�

�anŒpnC1.x/qn.0/�pn.x/qnC1.0/� anŒpnC1.z/pn.0/�pn.z/pnC1.0/�

D a2n ŒpnC1.x/pn.z/ � pn.x/pnC1.z//. pnC1.0/qn.0/� pn.0/qnC1.0/�

D Dn.x; z/Bn.0; 0/ D �Dn.x; z/: ut

5.6 Zeros of Orthogonal Polynomials

In this section, we derive a number of interesting results about zeros of theorthogonal polynomials .

Proposition 5.28 Suppose that p 2 RŒx� has degree m 2 N and h p.x/; xjis D 0

for j 2 N0; j � m � 2: Then the polynomial p.x/ has m distinct real zeros.

5.6 Zeros of Orthogonal Polynomials 113

More precisely, if � is a solution of the moment problem for s and J is a closedinterval containing supp�, then all zeros of p lie in the interior of J.

Proof First we recall that by Theorem 3.8 there exists a solution � of the momentproblem for s. Let �1; : : : ; �k denote the distinct real points in the interior of J,where p changes sign, and put r.x/ D .x � �1/ : : : .x � �k/. If there is no such�j, we set rD1. Then the polynomial r.x/p.x/ does not change sign on J. Henceq.x/ WD �r.x/p.x/ � 0 on J for �D1 or �D� 1.

We shall prove that k D m. Assume to the contrary that k < m.If k D m � 1, then p is a real polynomial of degree m which has m � 1 distinct

real zeros. The latter is only possible if k D m, which is a contradiction. Hencek � m� 2. But then h p; ris D 0 by assumption, so that

ZJq.x/ d�.x/ D �

ZJp.x/r.x/ d�.x/ D �h p; ris D 0:

Since q.x/ � 0 on J, this implies that� has a finite support. But then s is not positivedefinite, which contradicts our standing assumption.

This proves that k D m. Thus p has m distinct real zeros �1; : : : ; �m: utIn particular, Proposition 5.28 applies orthogonal polynomials and yields

Corollary 5.29 If .Rn/n2N0 is an OPS, then Rn.x/ has n distinct real zeros.

Clearly, Corollary 5.29 holds for the polynomials pn. As noted in Sect. 5.4, thepolynomialsepn.x/ D s�1=2

0 a0qnC1.x/, n 2 N0, are orthonormal polynomials for thepositive definite sequence obtained from the shifted Jacobi matrix (5.32). Therefore,Corollary 5.29 applies toepn, hence also to qnC1, and we have the following

Corollary 5.30 For n 2 N, the polynomials pn.x/ and qnC1.x/ have n distinct realzeros.

The following corollary plays an essential role in the proof of Lemma 7.1 below.

Corollary 5.31 Let n 2 N0 and z; z0 2 C. If jIm zj � jIm z0j, then

jpn.z/j � jpn.z0/j and jqn.z/j � jqn.z0/j: (5.59)

Proof Let r be a polynomial of degree n 2 N which has n distinct real zeros, sayx1; : : : ; xn. Then r.z/ D c.z�x1/ : : : .z�xn/ for some c 2 C. Therefore it is easilyseen that jr.z/j � jr.z0/j when jIm zj � jIm z0j. By Corollary 5.30 this applies topn and qnC1 for n 2 N and gives (5.59). For the constant polynomials p0; q0; q1 theassertion (5.59) is trivial. ut

Let us denote the m zeros �.m/j of the polynomial pm in increasing order

�.m/1 < �

.m/2 < � � � < �.m/m :


Proposition 5.32 The zeros of pn.x/ and pnC1.x/ interlace strictly, that is,

�.nC1/1 < �

.n/1 < �

.nC1/2 < � � � < �.n/n < �

.nC1/nC1 ; n 2 N: (5.60)

Proof Since p0.x/ D 1, it follows from Eq. (5.51) that

p0nC1.x/pn.x/� p0

n.x/pnC1.x/ > 0 for x 2 R:

Setting x D �.nC1/j therein we get

p0nC1.�

.nC1/j /pn.�

.nC1/j / > 0 for j D 1; : : : ; nC 1: (5.61)

Since each zero �.nC1/j is simple and pnC1 has a positive leading coefficient,

we conclude that sign p0nC1.�

.nC1/j / D .�1/nC1�j: Therefore, sign pn.�

.nC1/j / D

.�1/nC1�j by (5.61). Thus, by the mean value theorem, pn.x/ has at least one zeroin each of the n intervals .�.nC1/

j ; �.nC1/jC1 /. Since pn has only n zeros, we conclude

that there is precisely one zero in each of these intervals. utCorollary 5.33 The two limits ˛s WD limn!1 �

.n/1 and ˇs WD limn!1 �

.n/n exist

in R [ f�1g [ fC1g.Proof By (5.60), the sequence .�.n/1 / is decreasing and the sequence .�.n/n / isincreasing. utProposition 5.34 The zeros of qnC1.x/ and pnC1.x/ strictly interlace, that is, if .nC1/1 <

.nC1/2 < � � � < .nC1/

n are the zeros of qnC1, then

�.nC1/1 <

.nC1/1 < �

.nC1/2 < � � � < �.nC1/

n < .nC1/n < �

.nC1/nC1 ; n 2 N:

Proof The proof is similar to the proof of Proposition 5.32. Setting x D �.nC1/j in

formula (5.52) of Corollary 5.25 we obtain

pn.�.nC1/j /qnC1.�.nC1/

j / D a�1n > 0: (5.62)

By (5.60), pn has precisely one zero in each of the open intervals .�.nC1/j ; �

.nC1/jC1 / and

this zero is simple. Hence pn has different signs at the end points and so does qnC1by (5.62). Therefore, qnC1 has at least one zero in .�.nC1/

j ; �.nC1/jC1 / for j D 1; : : : ; n.

Since qnC1 has n zeros, this gives the assertion. utSince Pn D

pDn=Dn�1 pn, the preceding assertions hold for the monic

polynomials Pn as well. In particular, the zeros of Pn and PnC1 strictly interlace.The next proposition says that two arbitrary interlacing finite sequences are zero

sequences of some monic orthogonal polynomials.

5.7 Symmetric Moment Problems 115

Proposition 5.35 Given real numbers �1; : : : ; �mC1; 1; : : : ; m satisfying

�1 < 1 < �2 < � � � < m < �mC1; (5.63)

there exists a positive definite sequence s D .sn/n2N0 such that s0 D 1 and a monicOPS .Pn/n2N0 for s such that

Pm.x/ D .x � 1/ � � � .x � m/ and PmC1.x/ D .x � �1/ � � � .x � �mC1/:(5.64)

Proof By Favard’s Theorem 5.10, it suffices to show that there exist real sequences.˛n/n2N0 and .ˇn/n2N0 such that ˛n > 0 for all n and a sequence .Pn/n2N0 of monicreal polynomials such that

PnC1.x/ D .x � ˇn/Pn.x/ � ˛nPn�1.x/; n 2 N0; (5.65)

where P�1.x/ D 0, P0.x/ D 1, and Pm and PmC1 are given by (5.64).Since deg .PmC1 � xPm/ � m, there is unique real number ˇmC1 such that

Rm�1.x/ WD PmC1.x/� .x � ˇm/Pm.x/ (5.66)

has degree at most m � 1. From the assumption (5.63) it follows that Rm�1 changessigns at the zeros of Pm. Hence there is at least one real zero of Rm�1 betweentwo zeros of Pm. Since deg Rm�1 � m � 1, there is precisely one zero of Rm�1between two zeros of Pm. Since PmC1. m/ < 0 by (5.63), we have Rm�1. m/ < 0

by (5.66), so the leading coeefficient of Rm�1 is negative. Hence we can writeRm�1 D �˛mPm�1 with ˛m > 0 for some unique monic real polynomialPm�1. Then,by construction, (5.65) is satisfied for n D m and the zeros of Pm�1 and Pm strictlyinterlace. Therefore, we can continue by induction and construct polynomialsPm�2; : : : ;P1;P0 D 1 such that (5.65) is fulfilled for n D m� 1;m� 2; : : : ; 0:

To construct the polynomials Pn for n � m C 2 it suffices to choose numbers˛n > 0 and ˇn 2 R and define Pn inductively by (5.65). utCorollary 5.36 For each monic polynomial P of degree m with m distinct real zerosthere exists a monic OPS .Pn/n2N0 for some positive definite sequence with s0 D 1

such that P D Pm.

5.7 Symmetric Moment Problems

As throughout, s D .sn/n2N0 is a real positive definite sequence.

Definition 5.37 We say that s is symmetric if s2nC1 D 0 for all n 2 N0. A measure� on R is called symmetric if �.M/ D �.�M/ for all measurable sets M.


Clearly, s is symmetric if and only if

Ls. p.x// D Ls. p.�x// for p 2 CŒx�: (5.67)

Further, if a measure � 2 MC.R/ is symmetric, its odd moments vanish, henceits moment sequence is symmetric. Conversely, each symmetric positive definitesequence s has a symmetric measure� 2Ms. (Indeed, if � 2Ms, then the measure� defined by �.M/ D 1

2.�.M/C �.�M// is symmetric and belongs to Ms.)

Proposition 5.38 Let s be a symmetric positive definite sequence. Then

pn.�x/ D .�1/npn.x/; qn.x/ D .�1/nC1qn.x/; bn D 0 for n 2 N0:

(5.68)

The operator V defined by .Vp/.x/ D p.�x/, p 2 CŒx�, extends to a self-adjointunitary operator on the Hilbert spaceHs and VXV�1 D �X.Proof Let p; q 2 CŒx�. Using Eq. (5.67) we obtain

hVp; qis D Ls. p.�x/q.x// D Ls. p.x/q.�x// D h p;Vqis;hVp;Vqis D Ls. p.�x/q.�x// D Ls. p.x/q.x// D h p; qis;

so V is a symmetric isometric linear operator on the dense subspace CŒx�. Hence Vextends by continuity to a self-adjoint unitary on Hs. Obviously, VXV�1 D �X.

Set Qpn WD .�1/nV. pn/. Since pn has degree n, Qpn has a positive leading coeffient.Since V is unitary, .Qpn/n2N0 is an orthonormal basis of .CŒx�; h�; �is/. Therefore, bythe uniqueness assertion of Proposition 5.1, Qpn D pn, so that pn.�x/ D .�1/npn.x/.

Using Corollary 5.18 and Eq. (5.67) we derive for n 2 N0,

qn.�x/ D Ls;y

�pn.y/� pn.�x/

y � .�x/�D �.�1/nLs;y

�pn.�y/ � pn.x/

�y � x

�

D .�1/nC1Ls;y�pn.y/� pn.x/

y � x

�D .�1/nC1qn.x/:

Finally, we prove that bn D 0. By (5.9) we have the three term recurrence relation

xpn.x/ D anpnC1.x/C bnpn.x/C an�1pn�1.x/:

Replacing x by �x, using that pk.�x/ D .�1/kpk.x/; and dividing by .�1/n we get

�xpn.x/ D �anpnC1.x/C bnpn.x/ � an�1pn�1.x/:

Adding both equations gives 2bnpn D 0. Hence bn D 0. ut

5.8 Exercises 117

5.8 Exercises

1. Let s be a positive definite real sequence and set Qs WD cs, where c > 0. Provethe following for n 2 N0:

a. LQs D cLs and Dn.Qs/ D cnC1Dn.s/:b. Qpn D c�1=2pn , Qqn D c1=2qn, and QPn D Pn, QQn D Qn.c. Qan D an and Qbn D bn:d. QAn.x; z/ D cAn.x; z/, QBn.x; z/ D Bn.x; z/, QCn.x; z/ D Cn.x; z/, QDn.x; z/ D

c�1Dn.x; z/.

Here all quantities with a tilde refer to the positive definite sequence Qs.Hints: Use (5.3) for pn, Proposition 5.18 for qn; and Proposition 5.6 for an; bn.

2. Let T be a symmetric linear operator with domain D.T/ on a Hilbert space suchthat TD.T/ D.T/ and let e 2 D.T/. Define sn D hTne; ei for n 2 N0.

a. Show that s D .sn/n2N0 is a positive semidefinite sequence.b. Show that there exists a positive Radon measure � 2M.R/ such that � 2

Ms.c. Show that s is positive definite if and only if the span of vectors Tne; n 2 N0;

is infinite-dimensional.

3. What is the norm of the monic polynomial Pn.z/?4. Let s be positive real sequence. Prove the following:

a. If pn.�x/ D .�1/npn.x/ for all n 2 N0, then s is symmetric.b. If bn D 0 for all n 2 N0, then s is symmetric.

In the remaining exercises we develop important examples of orthogonal poly-nomials. Detailed treatments can be found in standard books such as [Chi1] or [Is].

We begin with the Chebyshev polynomials Tn.

5. Show that for each n 2 N0 there is a unique polynomial Tn 2 RŒx� such thatTn.x/ D cos.n arccos x/; x 2 R, or equivalently, Tn.cos �/ D cos.n�/; � 2 R.

6. Show that the polynomials Tn are uniquely defined by the recurrence relationTnC1.x/ D 2xTn.x/� Tn�1.x/; n 2 N; with initial data T0.x/ D 1;T1.x/ D x:

7. Show that the leading coefficient of Tn is 2n�1 for n 2 N0.8. Set s2n D .2n/Š

22n.nŠ/2and s2nC1 D 0 for n 2 N0. Let � be the probability measure

on Œ�1; 1� defined by d�.x/ WD ��1.1 � x2/�1=2dx. Show that s D .sk/k2N0 isa moment sequence with representing measure �.Hint: Use [RW, p. 174 and p. 274].

9. Show thatR 1

�1 Tk.x/Tl.x/d�.x/ D 0 if k ¤ l,R 1

�1 Tk.x/2d�.x/ D 1

2if k 2 N

andR 1

�1 T0.x/2d� D 1, that is, the orthonormal polynomials pn of the moment

sequence s are p0 D T0 and pn Dp2Tn for n 2 N.


Now we turn to the Hermite polynomials Hn. Define polynomials Hn by

HnC1.x/ D 2xHn.x/� 2nHn�1.x/; n 2 N; and H0.x/ D 1;H1.x/ D 2x:

10. Let s0 D 1, s2nC1 D 0; s2n D 2�n.2n� 1/ŠŠ WD 2�n � 1 � 3 � � � .2n� 1/ for n 2 N.Let � be the probability measure on R given by d� D 1p

�e�x2dx. Show that

s D .sk/k2N0 is a moment sequence with representing measure �.11. Show that � is the unique representing measure of s.12. Prove that

RRHk.x/Hl.x/d�.x/ D 2kkŠıkl for k; l 2 N0; that is,

. 1p2nnŠ

Hn.X//n2N0 is the sequence of orthonormal polynomials associatedwith s.

13. Show that Hn.x/ D .�1/nex2 � ddx

�ne�x2 and d

dxHnC1.x/ D 2.nC 1/Hn.x/ forn 2 N0:

Next we treat the Laguerre polynomials L˛n .x/, where ˛ > �1.Define a sequence of polynomials L˛n by L˛0 D 1;L˛1 .x/ D �xC 1C ˛, and

L˛nC1.x/ D�xC 2nC 1C ˛

nC 1 L˛n .x/�nC ˛nC 1 ; n 2 N:

14. Let sn D � .nC1C˛/� .˛C1/ for 2 N0 and let � be the probability measure on Œ0;C1/

defined by d� D .� .˛ C 1//�1x˛e�xdx: Show that s is a moment sequencewith representing measure �.

15. Show that � is the only representing measure for s.16. Show that

R10

L˛k .x/L˛l .x/d� D .˛C1/:::.˛Ck�1/

kŠ ıkl:

17. Show that L˛n .x/ D 1nŠx

�˛ex�ddx

�n.xnC˛e�x/ for n 2 N0.

18. Show that the leading coefficient of L˛n .x/ is .�1/nnŠ :

Finally, we develop the Legendre polynomials Rn.Define a sequence of polynomials Rn by

.nC 1/PnC1.x/ D .2nC 1/xRn.x/ � nRn�1.x/; n 2 N; and R0.x/ D 1;R1.x/ D x:

19. Verify that the Lebesgue measure on the interval Œ�1; 1� has the momentsequence s D .sk/k2N0 , where s2n D 2n

2nC1 and s2nC1 D 0.

20. Show thatR 1

�1 Rk.x/Rl.x/dx D 22kC1 ıkl for k; l 2 N0, that is,

pn.x/ DqkC 1

2Rn.x/, n 2 N; are the orthonormal polynomials for s.

21. Show that Rn.x/ D .�1/n2nnŠ

�ddx

�n..1� x2/n/ for n 2 N0.

5.9 Notes 119

5.9 Notes

The study of orthogonal polynomials is a large classical subject which is treatedin many books such as [Sz], [Chi1], [DX], [Is], [Sim2], [Sim3]. We do not makean attempt to discuss the history of this subject and mention only a few highlights.A number of formulas such as (5.6) and (5.7) go back to E. Heine (1878) [He].Favard’s theorem is in [Fv], see [MA] for a discussion of the history of this result.Proposition 5.35 was proved by B. Wendroff [Wen].

Chapter 6The Operator-Theoretic Approachto the Hamburger Moment Problem

In this chapter we begin the study of moment problems using self-adjoint operatorsand self-adjoint extensions on Hilbert spaces. The operator-theoretic approach is apowerful tool and it will be used in the next two chapters as well.

In Sect. 6.1, solutions of the Hamburger moment problem are related to spectralmeasures of self-adjoint extensions of the multiplication operator X (Theorem 6.1).In Sect. 6.3 we show that the moment problem is determinate if and only if theoperator X is essentially self-adjoint (Theorem 6.10) and we characterize vonNeumann solutions (Theorem 6.13). The multiplication operator X is unitarilyequivalent to the Jacobi operator T. In Sect. 6.2 the adjoint of the operator T isanalyzed, while in Sect. 6.4 various determinacy conditions (Theorem 6.16 andCorollary 6.19) are derived. In Sect. 6.5, all self-adjoint extensions of the symmetricoperator T on the Hilbert space l2.N0/ are described (Theorem 6.23). In Sect. 6.6we prove Markov’s theorem (Theorem 6.29) for determinate moment sequences.Section 6.7 gives a short disgression into continued fractions.

Throughout this chapter, s D .sn/n2N0 is a positive definite real sequence.

6.1 Existence of Solutions of the HamburgerMoment Problem

In this section we rederive the existence theorem for the Hamburger momentproblem from the spectral theorem for self-adjoint operators. This is not only theshortest, but probably also the most elegant and natural approach to this result.

Theorem 6.1 Let s be a positive definite real sequence. Then the Hamburgermoment problem for s has a solution.


121

122 6 The Operator-Theoretic Approach to the Hamburger Moment Problem

Let A be a self-adjoint operator on a Hilbert space G such thatHs is a subspaceof G and X A: If EA denotes the spectral measure of A, then

�A.�/ D hEA.�/1; 1iG (6.1)

is a representing measure for s: Every representing measure for s is of this form.

Proof Suppose that A is a self-adjoint extension of X on G. By the spectral theorem(see Appendix A.7 or [Sm9]), A has a spectral measure EA. Since X A and hence.X/n An, the polynomial 1 is in the domain D.An/ and we have

ZR

xn d�A.x/ DZR

xndhEA.x/1; 1i D hAn1; 1i D h.X/n1; 1is D hxn; 1is D Ls.xn/ D sn

for n 2 N0. This shows that �A is a solution of the moment problem for s.That each solution is of the form (6.1) follows from Proposition 6.2 below.To prove that the moment problem for s has a solution it therefore suffices to show

that the symmetric operator X has a self-adjoint extension. Define .Jp/.x/ D p.x/,p 2 CŒx�. Then J is a conjugation (see (A.28)) on CŒx� (that is, J is antilinear, J2 D I,and hJp; Jqis D hq; pis for p; q 2 CŒx�) which commutes with X (that is, JXp D XJpfor p 2 CŒx�). Clearly, J extends by continuity to a conjugation on Hs. Hence, byProposition A.43, X has a self-adjoint extension on the Hilbert space Hs. utProposition 6.2 Let � be a representing measure for s. Then

h p; qis D h p; qiL2.R;�/ for p; q 2 CŒx�:

The inclusion CŒx� L2.R; �/ extends to a unitary operator of Hs on a closedsubspace of L2.R; �/: We identify Hs with this closed subspace via this unitarymapping. Then the operator A� on L2.R; �/ defined by

.A� f /.x/ D xf .x/ for f 2 D.A�/ WD f f 2 L2.R; �/ W xf .x/ 2 L2.R; �/g (6.2)

is a self-adjoint extension of the symmetric operator X and

�.�/ D �A�.�/ � hEA�.�/1; 1iL2.R;�/:

Proof From � 2MC.R/ it follows that CŒx� D.A�/. Obviously, X A�. Since� is a representing measure for s, we have Ls. f / D

Rf d� for f 2 CŒx�, so that

h p; qis D Ls. pq/ DZ

p.x/q.x/ d� D h p; qiL2.R;�/:

Hence the inclusion CŒx� L2.R; �/ can be extended by continuity to a unitaryembedding of Hs into G WD L2.R; �/.

6.2 The Adjoint of the Jacobi Operator 123

From Hilbert space operator theory it is known (see e.g. [Sm9, Example 5.2]) thatthe operator A� is self-adjoint and that the spectral projection EA�.M/ acts as themultiplication operator by the characteristic function �M of M on L2.R; �/. Hence�.M/ D R �M.x/d� D hEA�.�/1; 1iL2.R;�/. The latter proves that � D �A� . ut

The following corollary reformulates the second assertion of Theorem 6.1 interms of the Jacobi operator T associated with s:

Corollary 6.3 The representing measures of s are precisely the measures of theform

�B.�/ D s0hEB.�/e0; e0iF ; (6.3)

where B is a self-adjoint extension of T on a possibly larger Hilbert space F andEB is the spectral measure of B.

Proof Recall from Sect. 5.3 that X D U�1TU, where U is the unitary of Hs ontol2.N0/ defined by Upn D en, n 2 N0: The sets of self-adjoint extensions A of Xand B of T are in one-to-one correspondence by a unitary equivalence. It sufficesto extend U to a unitary, denoted again by U, of G onto F such that A D U�1BU.Then EA D U�1EBU. Since p0.x/ D s�1=2

0 , we have U1 D s1=20 e0 and hence

�A.�/ D hEA.�/1; 1iG D hU�1EB.�/U 1; 1iGD hEB.�/U1;U1iF D s0hEB.�/e0; e0iF D �B: ut

Note that the measure hEB.�/e0; e0iF in (6.3) is always a probability measure,since e0 is a unit vector.

By Proposition 6.2, for each representing measure � the canonical Hilbert spaceHs is a closed subspace of L2.R; �/ and X is a restriction of the self-adjoint operatorA� on L2.R; �/. We shall see in Sect. 7.4 that for most solutions in the indeterminatecase Hs is a proper subspace of L2.R; �/.

The following definition gives a name for those solutions obtained from self-adjoint extensions acting on the same Hilbert space Hs.

Definition 6.4 A measure� 2Ms is called a von Neumann solution of the momentproblem for s if CŒx� is dense in L2.R; �/, or equivalently, if the embedding of CŒx�into L2.R; �/ extends to a unitary operator of Hs onto L2.R; �/.

6.2 The Adjoint of the Jacobi Operator

Recall from Sect. 5.3 that the Jacobi operator T � TJ is the symmetric linearoperator with dense domain d in the Hilbert space l2.N0/ given by

.T�/n D an�nC1 C bn�n C an�1�n�1; n 2 N0; (6.4)


for .�n/ 2 d, where ��1 WD 0: The next proposition shows that the adjoint operatorT� is the “maximal operator” on l2.N0/ which acts by the same formula (6.4). Forthis we essentially use the Wronskian defined by (5.40) and Lemma 5.23.

Proposition 6.5 The adjoint operator T� is given by

T�� D T � for � 2 D.T�/ D f� 2 l2.N0/ W T � 2 l2.N0/g:

For �; ˇ 2 D.T�/, the limit W.�; ˇ/1 WD limn!1 W.�; ˇ/n exists and

hT��; ˇi � h�;T�ˇi D W.�; ˇ/1 : (6.5)

Proof Let � 2 l2.N0/ be such that T � 2 l2.N0/. A straightforward computationshows that hTˇ; �i D hˇ; T �i for ˇ 2 d. Therefore, � 2 D.T�/ and T�� D T � .

Conversely, let � 2 D.T�/ and n 2 N0. Using (6.4) (or (5.22)) we derive

hen;T��i D hTen; �i D an�nC1 C bn�n C an�1�n�1 D .T��/n;

so that T�� D T � . This proves the first assertion concerning T�.Further, by (5.41) we have

nXkD0

Œ.T �/kˇk � �k.T ˇ/k/� D W.�; ˇ/n:

Since �; ˇ; T �; T ˇ 2 l2.N0/, the limit n!1 in the preceding equality exists andwe obtain Eq. (6.5). utProposition 6.6 Suppose that z 2 C.

(i) N .T��zI/ D f0g if pz … l2.N0/ andN .T��zI/ D C�pz if pz 2 l2.N0/.(ii) If h 2 N .X��zI/ and hh; 1is D 0, then h D 0.Proof

(i) From Proposition 6.5 it follows that a sequence � is in N .T��zI/ if and only if� 2 l2.N0/, T � 2 l2.N0/ and the recurrence relation (5.29) holds for n 2 N0,where ��1 D 0. Since ��1 D 0, any solution � of (5.29) is uniquely determinedby the number �0, so we have � D �0pz. This implies the assertions.

(ii) Passing to the unitarily equivalent operator T the assertion says that �0 D 0 and� 2 N .T��zI/ imply � D 0. Since � D �0pz, this is indeed true. ut

Corollary 6.7 The symmetric operator T (or equivalently X) has deficiency indices.0; 0/ or .1; 1/. The operator T (or X) is essentially self-adjoint if and only if pz isnot in l2.N0/, or equivalently

P1nD0 jpn.z/j2 D1, for one (hence for all) z 2 CnR.

Proof Since pn.x/ 2 RŒx� and hence pn.z/ D pn.z/, we have pz 2 l2.N0/ if andonly if pz 2 l2.N0/ for z 2 C. Therefore, by Proposition 6.6(i), T has deficiencyindices .0; 0/ or .1; 1/ and T has deficiency indices .0; 0/ if and only if pz … l2.N0/

for some (then for all) z 2 CnR. ut

6.3 Determinacy of the Hamburger Moment Problem 125

6.3 Determinacy of the Hamburger Moment Problem

We begin with two technical lemmas which are used in the proofs of Theorems 6.10and 6.13 below.

Lemma 6.8 If A and B are different self-adjoint extensions of the multiplicationoperator X onHs, then h.A�zI/�11; 1is ¤ h.B�zI/�11; 1is for all z 2 CnR.

Proof Fix z 2 CnR and assume to the contrary that

h.A�zI/�11; 1is D h.B�zI/�11; 1is: (6.6)

Put f WD .A�zI/�11 � .B�zI/�11. Since X A and X B, we have A X� andB X�. Hence f 2 D.X�/ and

.X��zI/f D .X��z/.A�zI/�11 � .X��zI/.B�zI/�11 D 1 � 1 D 0;

so f 2 N .X��zI/. Since h f ; 1is D 0 by (6.6), Proposition 6.6(ii) yields f D 0.Set gWD.A�zI/�11. If g were in D.X /, then for h 2 N .X��zI/ we would get

0 D h.X� � zI/h; gis D hh; .X�zI/gis D hh; .A�zI/.A�zI/�11is D hh; 1is;

so h D 0; again by Proposition 6.6(ii). Thus, N .X��zI/ D f0g. This is acontradiction, since X has two different self-adjoint extensions and hence itsdeficiency indices are .1; 1/ by Proposition A.42 and Corollary 6.7. Hence g is notin D.X /.

Let S denote the restriction of A to D.X/ C C�g. Then S is symmetric, becauseA is self-adjoint. Since X, hence X, has deficiency indices .1; 1/ by Corollary 6.7and g … D.X/, S has deficiency indices .0; 0/. Therefore S is self-adjoint and henceS D A. (In fact, the operator S is closed, but this is not needed here.)

Since f D 0 and hence g D .B�zI/�11, the same reasoning with B in place of Ashows that S D B. Thus A D B, which contradicts our assumption and shows thatEq. (6.6) cannot hold. utLemma 6.9 Suppose that � 2 MC.R/ is a finite measure. Let fz.x/ denote thefunction 1

x�z from L2.R; �/ for z 2 CnR. Then Ez0 D Lin f. fz0/k; . fz0 /k W k 2 N0gfor z0 2 CnR and E D Lin f fz W z 2 CnRg are dense linear subspaces of L2.R; �/:Proof Let ' 2 L2.R; �/. Since�.R/ <1, there is a finite complex Radon measure�' on R given by d�' D 'd�. Its Stieltjes transform I�' .z/ WD

RR

1x�z d�'.x/ is

holomorphic on CnR: Using Lebesgue’s convergence theorem we obtain

.I�' /.k/.z/ D kŠ

ZR

1

.x � z/kC1d�'.x/ for k 2 N0: (6.7)


Let us begin with Ez0 and assume that '?Ez0 : Let k 2 N0. Then, by (6.7),

0 D kŠ h'; . fz0/kC1iL2.R;�/ D kŠZ

'.x/

.x � z0/kC1d�.x/ D .I�' /.k/.z0/:

Similarly, .I�' /.k/.z0/ D 0 for k 2 N0. Therefore, since I�' is holomorphic on CnR,

we conclude that I�' .z/ D 0 in the upper half plane and in the lower half-plane. Thatis, the Stieltjes transform I�' is zero on CnR. From Theorem A.13 it follows thatthe measure �' is zero.

We prove that ' D 0 �-a.e. on R. For " > 0 put M" WD fx 2 R W j'.x/j � "g:Then, since the measure �' is zero and the measure � is positive, we derive

0 DZM"

'.x/ d�' DZM"

j'.x/j2d� �ZM"

"2d�.x/ D "2�.M"/ � 0:

Hence �.M"/ D 0. Letting " ! C0, we get �.fx 2 R W '.x/ ¤ 0g/ D 0; that is,' D 0 �-a.e.. Thus, ' D 0 in L2.R; �/. This proves that Ez0 is dense in L2.R; �/.

Now assume that '?E: Then '?fz means that I�' .z/ D 0. Hence the Stieltjestransform I�' .z/ is zero on CnR, so that �' D 0 by Theorem A.13. As shown in thepreceding paragraph this implies 'D0. This proves that E is dense in L2.R; �/: ut

The following theorem gives an operator-theoretic answer to the uniquenessproblem for the Hamburger moment problem.

Theorem 6.10 The moment problem for a positive definite sequence s is determi-nate if and only if the multiplication operator X (or equivalently, the correspondingJacobi operator T) is essentially self-adjoint. If this holds and � is the uniquerepresenting measure for s, then CŒx� is dense in L2.R; �/, that is, Hs Š L2.R; �/,so � is a von Neumann solution of the moment problem for s.

Proof First assume that X is not essentially self-adjoint. Then, by Corollary 6.7,X has deficiency indices .1; 1/. Therefore, by Proposition A.42, X has at least twodifferent self-adjoint extensionsA and B on Hs. By Theorem 6.1,�A.�/DhEA.�/1; 1isand�B.�/DhEB.�/1; 1is are representing measures for s. If �A were equal to �B, thenfor z 2 CnR the functional calculus would yield

h.A�zI/�11; 1is DZ.x � z/�1d�A.x/ D

Z.x � z/�1d�B.x/ D h.B�zI/�11; 1is;

which contradicts Lemma 6.8. Thus �A ¤ �B and s is indeterminate.Suppose now that X is essentially self-adjoint. Fix z 2 CnR. Since X is

essentially self-adjoint, .X�zI/CŒx� is dense in Hs, again by Proposition A.42.Hence there exists a sequence .rn.x// of polynomials such that 1 D limn.x�z/rnin Hs. (The sequence .rn/ may depend on the number z, but it is crucial that itis independent of any representing measure.) Let � be an arbitrary representingmeasure for s. Since � is finite, the bounded function 1

x�z is in L2.R; �/\L1.R; �/.

6.3 Determinacy of the Hamburger Moment Problem 127

Using the equations Ls. p/ DRp d� for p 2 CŒx�; k1k2

L2.R;�/D s20, and the Hölder

inequality we derive

jI�.z/ � Ls.rn/j2 Dˇˇ Z

R

.x � z/�1 d�.x/�ZR

rn.x/ d�.x/

ˇˇ2

��Z

R

j.x � z/�1 � rn.x/j d�.x/�2

� k1k2L2.R;�/ZR

j.x�z/�1 � rn.x/j2 d�.x/ (6.8)

D s20

ZR

jx�zj�2j1 � .x�z/rn.x/j2 d�.x/

� s20 jIm zj�2ZR

j1� .x�z/rn.x/j2 d�.x/

D s20 jIm zj�2 k1 � .x�z/rn.x/k2s ! 0:

Therefore I�.z/ D limn Ls.rn/ is independent of the representing measure �. ByTheorem A.13 the values of the Stieltjes transform I� determine the measure�. Hence � is uniquely determined by s, that is, the moment problem for s isdeterminate.

The preceding inequalities, especially (6.8), imply that for z 2 CnR the functionfz.x/ D 1

x�z is in the closure of CŒx� in L2.R; �/. Since the span E of such functionsis dense in L2.R; �/ by Lemma 6.9, so is CŒx�. utCorollary 6.11 Suppose that � 2 MC.R/. Then the moment sequence s of � isdeterminate if and only if CŒx� is dense in L2.R; .1C x2/d�/.

Proof Without loss of generality we can assume that s is positive definite. Indeed,otherwise � has finite support (by Proposition 3.11), hence the image of CŒx�coincides with L2.R; .1C x2/d�/, and s is determinate by Corollary 4.2.

By Theorem 6.10, s is determinate if and only if X is essentially self-adjoint, orequivalently by Proposition A.42, .x C z/CŒx� is dense in Hs for z D ˙i. Recallthat Hs is a subspace of L2.R; �/ by Proposition 6.2 and Hs Š L2.R; �/ if sis determinate. Therefore, s is determinate if and only if .x C z/CŒx� is dense inL2.R; �/ for z D ˙i. It is easily checked that this holds if and only if CŒx� is densein L2.R; .1C x2/d�/. (It suffices to approximate all functions of Cc.R/.) ut

By Proposition A.42, the operator X is essentially self-adjoint if .x ˙ i/CŒx� isdense in Hs. In the present situation we have the following stronger criterion.

Corollary 6.12 If there exist a number z0 2 CnR and a sequence .rn.x//n2N ofpolynomials rn 2 CŒx� such that

1 D limn!1.x � z0/rn.x/ in Hs;

then the moment problem for s is determinate.


Proof Fix p 2 CŒx�. Since z0 is a zero of the polynomial p.x/�p.z0/, there is apolynomial q 2 CŒx� such that p.x/�p.z0/ D .x�z0/q.x/. Then

.x�z0/.qCp.z0/rn/ D p�p.z0/C p.z0/.x�z0/rn ! p � p.z0/C p.z0/1 D p:

Therefore, because CŒx� is dense in Hs, .X�z0I/CŒx� is dense in Hs. Since X hasequal deficiency indices by Corollary 6.7, X is essentially self-adjoint. Hence s isdeterminate by Theorem 6.10. utTheorem 6.13 For a measure � 2Ms the following statements are equivalent:

(i) � is von Neumann solution.(ii) fz.x/ WD 1

x�z is in the closure of CŒx� in L2.R; �/ for all z 2 CnR.

(iii) fz0 .x/ WD 1x�z0

is in the closure of CŒx� in L2.R; �/ for one z0 2 CnR.

Proof(i)!(ii) That � is a von Neumann solution means that CŒx� is dense in L2.R; �/.

Hence the function fz 2 L2.R; �/ is in the closure of CŒx�.

(ii)!(iii) is trivial.

(iii)!(i) Set b WD Im z0. Let G denote the closure of CŒx� in L2.R; �/. We firstprove by induction on k that f kz0 2 G for all k 2 N. For k D 1 this is true byassumption. Suppose that f kz0 2 G. Then there is a sequence . pn/n2N of polynomialssuch that pn ! f kz0 in L2.R; �/. We can write pn.x/ D pn.z0/ C .x�z0/qn.x/ withqn 2 CŒx�. Using that j fz0.x/j � jbj�1 on R we derive

k f kC1z0 .x/ � pn.z0/.x�z0/�1�qn.x/kL2.�/D k fz0 .x/. f kz0.x/ � pn.z0/�.x � z0/qn.x//kL2.�/D k fz0 .x/. f kz0.x/ � pn.x//kL2.�/� jbj�1k f kz0 .x/ � pn.x/kL2.�/ ! 0 as n!1:

Since fz0 2 G by assumption and hence pn.z0/.x�z0/�1�qn 2 G, this shows thatf kC1z0 2 G; which completes the induction proof.

Clearly, f kz0 2 G implies that f kz0 2 G. Hence the vector space Ez0 from Lemma 6.9is contained in G. Since Ez0 is dense in L2.R; �/ by Lemma 6.9, CŒx� is dense inL2.R; �/. Hence � is a von Neumann solution. ut

We close this section by developing an operator-theoretic construction of inde-terminate moment sequences.

Example 6.14 For ˛ 2 R, let S˛ denote the operator �i ddx with dense domain

D.S˛/ D f f 2 ACŒ0; 2�� W f 0 2 L2.0; 2�/; f .2�/ D ei ˛2� f .0/ g

6.4 Determinacy Criteria Based on the Jacobi Operator 129

of L2.0; 2�/ with Lebesgue measure. Here ACŒ0; 2�� are the absolutely continuousfunctions on Œ0; 2��. (Recall that a function f on Œ0; 2�� is called absolutelycontinuous if there exists a function h 2 L1.0; 2�/ such that f .x/ D f .a/C R x

a f .t/dton Œ0; 2��; in this case Tf D �i h.) Then S˛ is a self-adjoint operator with spectrum�.S˛/ D f˛Ck W k 2 Zg, see e.g. [Sm9, p. 16, 34]. Each number ˛Ck is eigenvalueof multiplicity one with normalized eigenfunction '˛;k.x/ D 1p

2�ei.˛Ck/x and the

functions '˛;k; k 2 Z; form an orthonormal basis of the Hilbert space L2.0; 2�/:(The latter follows from the spectral theory of self-adjoint operators; it can also beverified directly. That the span of functions '˛;k; k 2 Z; is dense for ˛ 2 R followsat once from the denseness of functions '0;k; k 2 Z:)

We fix a function f 2 C10 .0; 2�/; f ¤ 0; and define

sn WD h.�i/nf .n/; f i; n 2 N0; and s D .sn/n2N0 :

Since f .n/.0/ D f .n/.2�/ D 0 for n 2 N0, f is in the domain of each power .S˛/n

and .S˛/nf D .�i/nf .n/ for ˛ 2 R and n 2 N0. We develop f with respect to theorthonormal basis f'˛;k W k 2 Zg and write f D P

k c˛;k'˛;k. For n 2 N0 we have.S˛/nf DPk c˛;n.˛ C k/n'˛;k and hence

sn D h.�i/nf .n/; f i D h.S˛/k; f i D�Xk2Z

c˛;k.˛ C k/n'˛;k;Xl2Z

c˛;l'˛;l

�

DXk2Z

c2˛;k.˛ C k/n DZ

xnd�˛; where �˛ WDXk2Z

c2˛;kı˛Ck:

Therefore, s is a moment sequence and �˛; ˛ 2 R; is a representing measure of s.Let ˛; ˇ 2 Œ0; 1/; ˛ ¤ ˇ: Since f ¤ 0, there is a k 2 Z such that c2˛;k ¤ 0. Hence

�˛.f˛C kg/ ¤ 0, while �ˇ.f˛C kg/ D 0: Thus �˛ ¤ �ˇ and s is indeterminate. ı

6.4 Determinacy Criteria Based on the Jacobi Operator

The following lemma is a main technical ingredient of the proof of Theorem 6.16.

Lemma 6.15 Suppose that c D .cn/; ' D .'n/; D . n/; D .n/, and � D .�n/are sequences from l2.N0/. If f D . fn/n2N0 is a complex sequence satisfying

fnC1 D cn C 'nnX

kD0k fk C n

nXkD0

�k fk for n 2 N0; (6.9)

then f 2 l2.N0/.


Proof Let " > 0. Since c; '; ; ; � 2 l2.N0/, there exists an l 2 N such that

mXnDl

jcnj2 C kk2mXnDl

j'nj2 C k�k2mXnDl

j nj2 � "

for all m � l. Suppose that m � l. Using the inequality

ˇˇ nXkD0

k fk

ˇˇ2 �

� nXkD0jkj2

�� nXkD0j fkj2

�� kk2

mXjD0j fkj2; m � n;

we derive

mXnDl

ˇˇ'n

nXkD0

k fk

ˇˇ2 �

mXnDl

j'nj2ˇˇ nXkD0

k fk

ˇˇ2 � kk2

mXjD0j fkj2

mXnDl

j'nj2 � "mXjD0j fkj2

and similarly

mXnDl

ˇˇ n

nXkD0

�k fk

ˇˇ2 � "

mXjD0j fkj2:

From the preceding two inequalities and Eq. (6.9) we therefore obtain

1

4

mXnDl

j fnC1j2 �mX

kDl

jckj2 CmX

nDl

ˇˇ'n

nXkD0

k fk

ˇˇ2 C

mXnDl

ˇˇ n

nXkD0

�k fk

ˇˇ2 � "C 2"

mXjD0

j fkj2

and hence

�14� 2"�

mXnDl

j fnC1j2 � "C 2"l�1XjD0j fkj2 for m � l:

Choosing " < 18

we conclude that f is in l2.N0/. utThe next theorem sharpens Corollary 6.7 and it is the main result of this section.

Theorem 6.16 For any positive definite sequence s the following are equivalent:

(i) The moment problem for s is indeterminate.(ii) The Jacobi operator T D TJ is not essentially self-adjoint.

(iii) pz 2 l2.N0/ for some (equivalently, for all) z 2 CnR.(iv) qz 2 l2.N0/ for some (equivalently, for all) z 2 CnR.(v) pz 2 l2.N0/ and qz 2 l2.N0/ for some (equivalently, for all) z 2 R.

(vi) pz 2 l2.N0/ and qz 2 l2.N0/ for some (equivalently, for all) z 2 C.

6.4 Determinacy Criteria Based on the Jacobi Operator 131

Proof First we note that (i)$(ii) by Theorem 6.10, (ii)$(iii) by Corollary 6.7,and (iii)$(iv) by Corollary 5.22. Further, (vi)!(iv) and (vi)!(v) are trivial. Theproof is complete once we have shown that (iii) and (iv) together imply (v) and that(v) implies (vi). For this it suffices to prove the following assertion:

If pz and qz are in l2.N0/ for some z 2 C, then this holds for all x 2 C.Indeed, suppose that pz and qz are in l2.N0/. Fix x 2 C. Set fn D pn.x/ and

cn D pnC1.z/; 'n D .x � z/qnC1.z/; n D .z � x/pnC1.z/; n D pn.z/; �n D qn.z/

for n 2 N0. Since pz; qz 2 l2.N0/, the sequences c D .cn/; ' D .'n/; D. n/; D .n/, and � D .�n/ are in l2.N0/. Further, from the identity (5.53) itfollows that

pnC1.x/ D pnC1.z/C .x�z/qnC1.z/nX

kD0pk.z/pk.x/C .z�x/pnC1.z/

nXkD0

qk.z/pk.x/:

This means that Eq. (6.9) holds. Therefore, all assumptions of Lemma 6.15 arefulfilled, so we obtain px D . fn/ 2 l2.N0/. Replacing (5.53) by (5.54) andproceeding verbatim in the same manner we derive that qx 2 l2.N0/. utRemark 6.17 The equivalence of conditions (i)–(iv) of Theorem 6.16 was obtainedby general operator-theoretic considerations. For the description of self-adjointextensions in Sect. 6.5 we need that p0 and q0 are in l2.N0/ if s is indeterminate. Thisis more tricky and follows from the implication (i)!(vi) and also from Lemma 7.1proved in the next chapter. ı

Combining Theorem 6.16 (i)!(vi), Proposition 6.5, and Lemma 5.16 we obtainthe following important corollary.

Corollary 6.18 If s is an indeterminate moment sequence, then for all (!) z 2 C thesequences pz D . pn.z//n2N0 and qz D .qn.z//n2N0 are in D.T�/,

T�pz D zpz and T�qz D s1=20 e0 C zqz : (6.10)

The next corollary contains another sufficient condition for determinacy.

Corollary 6.19 IfP1

nD0 a�1n D 1, then the Jacobi operator T is essentially self-

adjoint and the moment problem is determinate.

Proof Assume to the contrary that T is not essentially self-adjoint. Then, byTheorem 6.16, pz 2 l2.N0/ and qz 2 l2.N0/ for z 2 CnR. Since

a�1n D pn.z/qnC1.z/� pnC1.z/qn.z/;


by formula (5.52), we derive

1XnD1

a�1n D

1XnD1Œ pn.z/qnC1.z/ � pnC1.z/qn.z/�

�� 1X

nD1jpn.z/j2

�1=2� 1XnD1jqnC1.z/j2

�1=2C� 1X

nD1jpnC1.z/j2

�1=2� 1XnD1jqn.z/j2

�1=2

� kpzkl2.N0/ kqzkl2.N0/ C kqzkl2.N0/ kpzkl2.N0/ <1;which contradicts the assumption

P1nD0 a�1

n D 1. utRemark 6.20 Under additional assumptions a converse to Corollary 6.19 holds (see[Bz, Theorem 1.5, p. 507]): Suppose that the sequence .bn/n2N0 is bounded andan�1anC1 � a2n for n � n0 and some n0 2 N. If

P1nD0 a�1

n < 1, the Jacobioperator T is not essentially self-adjoint and the moment problem is indeterminate. ı

In Sect. 4.2 Carleman’s theorem 4.3 was derived from the Denjoy–Carlemantheorem 4.4 on quasi-analytic functions. Now we give an operator-theoretic proofof Theorem 4.3 which is based on Corollary 6.19.

Second Proof of Carleman’s Theorem 4.3(i) If s0 > 0 and .sn/ satisfies Carleman’scondition, then so does .sns�1

0 /. Hence we can assume that s0 D 1. Then,by Corollary 5.7, pn has the leading coefficient .a0 : : : an�1/�1. Therefore, sincehxk; pnis D 0 for 0 � k < n, we have

h.a0 : : : an�1/�1xn; pnis D h pn; pnis D 1:From the Cauchy–Schwarz inequality we obtain

1 � k.a0 : : : ; an�1/�1xnk2skpnk2s D .a0 : : : an�1/�2s2n;

so that

s� 12n

2n � .a0 : : : an�1/�1n ; n 2 N: (6.11)

Stirling’s formula (see e.g. [RW, p. 45]) yields . ne /n � nŠ and hence . 1nŠ /

1=n � en :

Using this fact and the arithmetic-geometric mean inequality we derive

�1

a0: : :

1

an�1

� 1n

D�1

nŠ

� 1n�1

a0

2

a1: : :

n

an�1

� 1n

� e

n

1

n

nXkD1

k

ak�1: (6.12)

Let us fix N 2 N. For k 2 N; k < N, we get

NXnDk

k

n2�

NXnDk

2k

n.nC 1/ � 2kNX

nDk

�1

n� 1

nC 1�D 2k

�1

k� 1

N C 1�< 2:

(6.13)

6.5 Self-Adjoint Extensions of the Jacobi Operator 133

Using first (6.11) and (6.12) and finally (6.13) it follows that

NXnD1

s� 12n

2n �NX

nD1

e

n2

nXkD1

k

ak�1D

NXkD1

NXnDk

e

n2k

ak�1�

NXkD1

e

ak�1

NXnDk

k

n2� 2e

N�1XkD0

1

ak:

Therefore, the assumptionP1

nD1 s�1=2n2n D 1 implies that

P1nD1 a�1

n D 1. Hences is determinate by Corollary 6.19. ut

The case when a Jacobi operator is bounded is clarified by the next proposition.In this case it is obvious that T is essentially self-adjoint, so s is determinate.

Proposition 6.21 The Jacobi operator T is bounded if and only if both sequencesa D .an/n2N0 and b D .bn/n2N0 are bounded.

Proof If T is bounded, the sequences a and b are bounded, since

jan�1j2 C jbnj2 C janC1j2 D kanenC1 C bnen C an�1en�1k2 D kTenk2 � kTk2:

Moreover, supn janj � kTk and supn jbnj � kTk.Conversely, assume that both sequences a and b are bounded, say janj � M and

jbnj � M for n 2 N0. Let � 2 d. Using the triangle inequality in l2.N0/ we derive

kT�k D �Xnjan�n�1 C bn�n C an�1�n�1j2/1=2

� �Xnjan�n�1j2

�1=2 C �Xnjbn�nj2/1=2 C

�Xnjan�n�1j2

�1=2 � 3Mk�k;that is, T is bounded and kTk � 3M: ut

6.5 Self-Adjoint Extensions of the Jacobi Operator

If a moment sequence is determinate, the Jacobi operator T is essentially self-adjointby Theorem 6.10 and hence its closure is the unique self-adjoint extension of T.

Throughout this section, s is an indeterminate moment sequence. By Theo-rem 6.10 and Corollary 6.7, T has deficiency indices .1; 1/. Our aim is to describeall self-adjoint extensions of the symmetric operator T on the Hilbert space Hs.

Recall from Corollary 6.18 that for each z 2 C the sequences pz; qz are in D.T�/and satisfy T�pz D zpz and T�qz D s1=20 e0 C zqz: In particular,

T�p0 D 0; T�q0 D s1=20 e0; and p0.z/ D s�1=20 ; q0.0/ D 0:

To simplify some computations we set p WD s1=20 p0 and q WD s�1=20 q0: Then

T�p D 0; T�q D e0; and p D .1; : : : /; q D .0; : : : /: (6.14)


These properties of p and q will play an essential role in what follows.Let PC denote the direct sum. Recall that T is the closure of the operator T.

Lemma 6.22 D.T�/ D D.T/ PCC � q0 PCC � p0:Proof Obviously,D.T/CC �q0CC �p0 D D.T/CC �qCC �p D.T�/. It sufficesto prove the converse inclusion of the latter.

Using (6.14) and the fact that the operator T is symmetric we compute

hT�.' C c0qC c1p/; C d0qC d1pi � h' C c0qC c1p;T�. C d0qC d1p/i

D hT' C c0e0; C d0qC d1pi � h' C c0qC c1p;T C d0e0iD h'; d0e0i C hc0e0; C d1pi � hc0e0; i � h' C c1p; d0e0iD '0d0 C c0 . 0 C d1 / � c0 0 � .'0 C c1/ d0 D c0 d1 � c1 d0 (6.15)

for arbitary '; 2 D.T/ and c0; c1; d0; d1 2 C.Since T has deficiency indices .1; 1/, we have dim D.T�/=D.T/ D 2 by formula

(A.27) in Appendix A.7. Therefore, to prove that D.T�/ D.T/C C � qC C � p itsuffices to show the vectors q and p are linearly independent modulo D.T/. Indeed,if c0qC c1p 2 D.T/, then

hT�.c0qC c1p/; d0qC d1pi D hc0qC c1p;T�.d0qCd1p/i

and hence c0 d1 � c1 d0 D 0 by (6.15) for arbitrary d0; d1 2 C. Therefore, c0 Dc1 D 0, so q and p are linearly independent modulo D.T/. utTheorem 6.23 The self-adjoint extensions of the Jacobi operator T on Hs Šl2.N0/ are precisely the operators Tt D T�dD.Tt/, t 2 R [ f1g, where

D.Tt/ D D.T/ PCC � .q0 C tp0/ for t 2 R; D.T1/ D D.T/ PCC � p0: (6.16)

Further, if the symmetric operator T is positive, so is the self-adjoint operator T1.

Proof Let A be a self-adjoint extension of T on the Hilbert space Hs. Because T Aand hence T� � A� D A, the operator A is completely described by its domainD.A/. Since T has deficiency indices .1; 1/ and hence dim D.T�/=D.T/ D 2, weconclude that dim D.A/=D.T/ D 1. Thus, up to complex multiples, there exists aunique 2 D.A/ which is not in D.T/. By Lemma 6.22 the vector can be chosento be of the form D c0q0 C c1p0 2 D.A/: Further, upon scaling we can write D s0qC tp with t 2 C or D p.

First we treat the case D s0q C tp. Let c0; d0; d1 2 C and 2 D.T/. Since' C c0 D ' C c0s0qC c0tp 2 D.A/ and A T�, it follows from (6.15) that

hA.' C c0/; C d0qC d1pi � h' C c0;T�. C d0qCd1p/i

D c0.s0 d1�t d0 /: (6.17)

6.5 Self-Adjoint Extensions of the Jacobi Operator 135

Let d0qCd1p be a nonzero vector in D.A/. Then d0qCd1p is a multiple of , so thats0d1 D d0t and d0 ¤ 0. Hence c0.s0d1 � t d0/ D c0s0d0. t � t/: Therefore it followsfrom (6.17) that the operator A is symmetric if and only if (6.17) vanishes for allc0 2 C, or equivalently, if t is real. Since D s0q C tp D s1=20 .q0 C tp0/ 2 D.A/,then we have A D Tt.

Now let D p. Then A D T1. If c1; d0; d1 2 C and 2 D.T/, then by (6.15),

hA.' C c1/; Cd0qCd1pi � h' C c1;T�. Cd0qCd1p/i D c1 d0: (6.18)

If d0qC d1p 2 D.T1/, then d0 D 0 and the right-hand side of (6.18) vanishes. Thisproves that A D T1 is symmetric.

We have shown that all operators Tt, t 2 R [ f1g, are symmetric. It remainsto prove that they are self-adjoint. Assume to the contrary that A D Tt is not self-adjoint. Then A ¤ A�. Since T A A� T� and dim D.T�/=D.A/ D 1 (bydim D.T�/=D.T/ D 2), we conclude that A� D T�. Hence the right-hand sideof Eq. (6.17) resp. (6.18) has to vanish. Since c0; d0; d1 2 C resp. c1; d0 2 C arearbitrary, this leads to a contradiction.

To prove the last assertion let ' 2 D.T/ and c 2 C. Since the symmetric operatorT is positive, so is its closure T. Using that T1p D T�p D 0 we obtain

hT1.' C cp/; ' C cpi D hT'; ' C cpi D hT'; 'i C ch';T�pi D hT'; 'i � 0:

This shows that T1 is positive. utFor t 2 R [ f1g we set �t.�/ WD s0hEt.�/e0; e0i, where Et denotes the spectral

measure of the self-adjoint operator Tt. The next lemma plays an essential role inthe proof of Theorem 7.6 below. It shows how the parameter t 2 R can be recoveredfrom the measure �t and from the operator Tt.

Lemma 6.24 For t 2 R, the operator Tt is invertible and

limy2R;y!0

I�t . yi/ D s0 limy2R;y!0

h.Tt�yiI/�1e0; e0i D s0hT�1t e0; e0i D t: (6.19)

Further, limy2R;y!0

jI�1. yi/j D C1:

Proof Let t 2 R. Since N .Tt/ N .T�/ D C � p by Proposition 6.6(i) and p …D.Tt/;we have N .Tt/Df0g, so the operator Tt is invertible. Recall that qCs�1

0 tp Ds�1=20 .q0 C tp0/ 2 D.Tt/: From T�p D 0 and T�q D e0 by (6.14) it follows thatTt.qC s�1

0 tp/ D T�.qC s�10 tp/ D e0. Thus e0 2 D.T�1

t / and T�1t e0 D qC s�1

0 tp,so that

s0hT�1t e0; e0i D hs0qC tp; e0i D t: (6.20)

Since e0 2 D.T�1t /, the function h.x/ D x�2 is �t-integrable. In particular,

�t.f0g/ D 0. We set hy.x/ D j.x�yi/�1 � x�1j2 for y 2 .�1; 1/: Then hy.x/ ! 0


�t–a.e. on R as y ! 0 and jhy.x/j � h.x/ for y 2 .�1; 1/: Therefore, Lebesgue’sdominated convergence theorem applies and using the functional calculus for theself-adjoint operator Tt we derive

k.Tt�yiI/�1e0 � T�1t e0k2 D

ZR

j.x � yi/�1 � x�1j2 dhEt.x/e0; e0i

D s�10

ZR

j.x � yi/�1 � x�1j2 d�t.x/! 0 as y! 0:

Therefore, limy!0

.Tt�yiI/�1e0 D T�1t e0. Combining this with (6.20) and using the

equality I�t . yi/ D s0h.Tt�yiI/�1e0; e0i; by the functional calculus, we get (6.19).

Since T1e0 D 0, �1.f0g/ > 0. Hence it follows from jI�1. yi/j �

�1.f0g/jyj�1 that jI�1. yi/j ! C1 as R 3 y! 0. ut

In the remaining part of this section we consider symmetric moment sequences.

Proposition 6.25 Suppose that the indeterminate moment sequence s is symmetric,that is, s2nC1 D 0 for n 2 N0. Let V be the self-adjoint unitary operator (seeProposition 5.38) of the Hilbert space Hs defined by V. p/.x/ D p.�x/, p 2 CŒx�:Then VT0V�1 D �T0 and VT1V�1 D �T1. Further, if t 2 R[f1g and VD.Tt/ D.Tt/, then t D 0 or t D 1.

Proof Since s is symmetric, pn.�x/ D .�1/npn.x/ and qn.x/ D .�1/nC1qn.x/ by(5.68). Therefore, p2kC1.0/ D q2k.0/ D 0 for all k 2 N0. Hence the Fourier seriesof p0 and q0 with respect to the orthonormal basis . pn.x//n2N0 of the Hilbert spacel2.N0/ Š Hs have the form

p Š1XkD0

p2k.0/p2k.x/ and q Š1XkD0

q2kC1.0/p2kC1.x/:

Hence Vp0 D p0 and Vq0 D �q0, so that V.q0C tp0/ D �q0C tp0. Thus, by (6.16),the relation VD.Tt/ D.Tt/ holds if and only if the parameter t is 0 or1:

Recall that VTV�1 D �T by Proposition 5.38. (As throughout, we identify X andT via the canonical unitary isomorphism of Hs and l2.N0/.) Hence VD.T/ D D.T/and VT�V�1 D �T�. As noted in the preceding paragraph, VD.Tt/ D D.Tt/ fort D 0;1. Therefore, since Tt T�, we conlude that VTtV�1 D �Tt for t D 0;1.

utCorollary 6.26 If s is a symmetric indeterminate moment sequence, then �0 and�1 are different symmetric (!) representing measures for s.

Proof Let t D 0 or t D 1. Then VTtV�1 D �Tt by Proposition 6.25. We define apositive Radon measure Q�t on R by d Q�t.�x/ D d�t.x/, x 2 R. Clearly, it follows

6.6 Markov’s Theorem 137

from that equality�Tt D VTtV�1 that .�Tt�zI/�1 D V.Tt�zI/�1V�1 for z 2 CnR.Using the relations V�1e0 D e0 and V D V� we derive

I Q�t.z/ D s0 h.�Tt � zI/�1e0; e0i D s0 hV.Tt � zI/�1V�1e0; e0iD s0 h.Tt � zI/�1e0; e0i D I�t .z/ for z 2 CnR;

that is, the Stieltjes transforms of Q�t and �t coincide. Hence Q�t D �t byTheorem A.13. This means that �t is symmetric.

To prove that �0 ¤ �1 we assume to the contrary that �0 D �1. Then we haveh.T0�zI/�1e0; e0i D h.T1�zI/�1e0; e0i. This contradicts Lemma 6.8, since T0 andT1 are different self-adjoint extensions of T. (Note that Lemma 6.8 was formulatedfor the operator X, but X is unitarily equivalent to T.) ut

6.6 Markov’s Theorem

In this section s D .sn/n2N0 is a positive definite sequence and a D .an/n2N0 andb D .bn/n2N0 are the sequences in the corresponding Jacobi matrix J.

Let Jn be the operator on Cn defined by the truncated Jacobi matrix

Jn D

0BBBBBBB@

b0 a0 0 : : : 0 0 0

a0 b1 a1 : : : 0 0 0

0 a1 b2 : : : 0 0 0

: : : : : : : : : : : : : : : : : : : : :

0 0 0 : : : an�3 bn�2 an�20 0 0 : : : 0 an�2 bn�1

1CCCCCCCA; n 2 N: (6.21)

This matrix and Lemmas 6.27 and 6.28 will be used later in Sect. 8.4 as well.The next lemma gives another description of the monic polynomial Pn and it

shows that eigenvalues and eigenvectors of the self-adjoint operator Jn can be nicelyexpressed in terms of zeros of the orthogonal polynomials pk.

Lemma 6.27

(i) Pn.z/ D det.zIn � Jn/ for n 2 N; z 2 C:

(ii) Let �1; : : : ; �n be the zeros of pn and y. j/ D . p0.�j/; : : : ; pn�1.�j// 2 Cn. Thenwe have Jny. j/ D �jy. j/ for j D 0; : : : ; n � 1 and n 2 N.

Proof

(i) By developing det.zInC1�JnC1/ after the last row we obtain

det.zInC1�JnC1/ D .z � bn/ det.zIn�Jn/� a2n�1 det.zIn�1�Jn�1/ (6.22)


for n 2 N. Note that (6.22) is also valid for n D 1 by setting det.zI0�J0/ WD 1.From (6.22) and Proposition 5.9 it follows that the sequence of polynomi-als det.zIn�Jn/ satisfy the same reccurence relation (5.14) and intial datadet.zI0�J0/ D P0.x/ and det.zI1�J1/ D z � b0 D P1.z/ as Pn. HencePn.z/ D det.zIn�Jn/ for n 2 N:

(ii) For n D 1 the assertion is easily checked, so we can assume that n � 2.Consider an equation Jny D zy, where y D . y0; : : : ; yn�1/ 2 Cn, y ¤ 0, and z 2C. For the first n�1 components this equation means that y0; : : : ; yn�1 satisfythe same recurrence relations (5.29) as the polynomials p0.z/; : : : ; pn�1.z/ dowith y�1Dp�1.z/ WD 0. Hence yk D cpk.z/ for k D 0; : : : ; n � 1 and somenonzero c 2 C. Inserting this into the n-th component of Jny D zy yields

zpn�1.z/ D an�2pn�2.z/C bn�1pn�1.z/:

On the other hand, by the relation (5.29) we have

zpn�1.z/ D an�1pn.z/CCbn�1pn�1.z/C an�2pn�2.z/:

Since an�1 ¤ 0; it follows that Jny D zy holds if and only if pn.z/ D 0. Hencethe eigenvalues of Jn are precisely the zeros �j of pn and the correspondingeigenvectors are the vectors y. j/. ut

Lemma 6.28 Let Kn be the matrix which is obtained from Jn by removing the firstrow and the first column. Then

det .zI � Kn/ D Qn.z/ for n 2 N; n � 2; z 2 C; (6.23)

h.Jn � zI/�1e0; e0i D �Qn.z/

Pn.z/for z 2 .Jn/: (6.24)

Proof It suffices to prove the assertion for n � 3. Consider an equation Kny D zyfor y D . y1; : : : ; yn�1/ 2 Cn�1, y ¤ 0, and z 2 C. Recall that Kn is the.n�1/�.n�1/matrix in left upper corner of the shifted Jacobi matrix QJ from (5.32).Therefore, setting y0 WD 0 and using that q0 D 0, the first n�2 components ofthe equation Kny D zy coincide with the recurrence relation for the polynomialsq1.z/; : : : ; qn�1.z/; see Sect. 5.4. Hence yj D cqj.z/ for j D 1; : : : ; n � 1 and somec 2 C; c ¤ 0: For the .n�1/-th component it follows in a similar manner as inthe proof of Lemma 6.27 that z is an eigenvalue of Kn if and only if qn.z/ D 0, orequivalently, Qn.z/ D 0. Thus, det .zI � Kn/ and Qn.z/ are monic polynomials ofdegree n� 1 having the same zeros. Hence these polynomials coincide. This proves(6.23).

Finally, we prove (6.24). Let z 2 .Jn/. Setting f D . f0; � � � ; fn/ WD .Jn�zI/�1e0,to compute f0 we apply Cramer’s rule and get

6.6 Markov’s Theorem 139

h.Jn � zI/�1e0; e0i D f0 D

ˇˇˇˇˇˇ

1 a0 0 : : : 0 0

0 b1 � z a1 : : : 0 0

: : : : : : : : : : : : : : : : : :

0 0 0 : : : bn�1 � z an�10 0 0 : : : an�1 bn � z

ˇˇˇˇˇˇ

det .Jn � zI/

D det .Kn � zI/

det .Jn � zI/D �det .zI � Kn/

det .zI�Jn/ D �Qn.z/

Pn.z/:

Here for the last equality we used (6.23) and Lemma 6.27(i). utEquation (6.25) in the following result is called Markov’s theorem.In particular, if supp� is bounded, or equivalently, if the operator T Š X is

bounded, then s is determinate and the equality (6.25) holds.

Theorem 6.29 Suppose that s is a positive definite determinate moment sequence.If � is the representation measure for s and R is an interval containing supp�, then

� limn!1

qn.z/

pn.z/D �s0 lim

n!1Qn.z/

Pn.z/DZR

d�.x/

x � zfor z 2 CnR: (6.25)

Proof The first equality in (6.25) follows at once from (5.36). To prove the secondequality we extend Jn to a finite rank operator QJn on l2.N0/ by setting

QJn WD�Jn 00 0

�on l2.N0/ D Cn ˚ l2.Nn/;

where Nn D fk 2 N W k � ng. Fix a number z 2 CnR.Because s is determinate, T Š X is essentially self-adjoint by Theorem 6.10 and

Hs Š L2.R; �/. Since the multiplication operator A� by the variable x in L2.R; �/(see Proposition 6.2) is a self-adjoint extension of X and so of Mx, we have A� DMx. Therefore, the spectrum of the operator T Š Mx D A� is the support of � andso a subset of R. Hence, since z 2 CnR, .T � zI/d is dense in l2.N0/. Further, sincePn.z/ D det .zI�Jn/ by Lemma 6.27(i), the spectrum of Jn, hence ofeJn, consistsof the zeros of Pn, so it is also contained in R by Corollary 5.28. Thus, z is in theresolvent sets of eJn and T.

Let ' D .'0; : : : ; 'k; 0; : : : / 2 d. If n � k, then we have .T � zI/' 2 CnC1 and.eJn � zI/' D .T � zI/', so that

�.eJn � zI/�1 � .T � zI/�1

�.T � zI/'

D .eJn � zI/�1.eJn � zI/' � .T � zI/�1.T � zI/' D ' � ' D 0:

Hence, since k.eJn � zI/�1k � .dist.z;R//�1 and .T � zI/d is dense, it follows that


limn!1.

eJn � zI/�1 D .T � zI/�1 for 2 l2.N0/: (6.26)

Since T is self-adjoint, we have �.�/ D s0hET.�/e0; e0i by formula (6.3) inCorollary 6.3. Therefore, applying (6.24) and (6.26) with D e0, we derive

� s0 limn!1

Qn.z/

Pn.z/D s0 lim

n!1 h.Jn � zI/�1e0; e0i D s0 limn!1 h.eJn � zI/�1e0; e0i

D s0 h.T � zI/�1e0; e0i D s0

ZR

.x � z/�1dhET.x/e0; e0i DZR

.x � z/�1d�.x/;

which proves the second and main equality of (6.25). ut

6.7 Continued Fractions

We begin with general continued fractions. Let .˛n/n2N and .ˇn/n2N0 be complexsequences. An (infinite) continued fraction is a formal expression

ˇ0 C ˛1

ˇ1 C ˛2

ˇ2C:::C ˛n

ˇnC

:::

: (6.27)

Just as in case of infinite series we want to associate a number to this expression.For this reason we set

Cn D ˇ0 C ˛1

ˇ1 C ˛2

ˇ2C:::C ˛n

ˇn

: (6.28)

Note that it may happen that Cn is not defined if some denominator is zero. To savespace the expressions (6.27) and (6.28) are written as

ˇ0 C ˛1jjˇ1 C

˛2jjˇ2 C � � � C

˛njjˇn C : : : ;

Cn D ˇ0 C ˛1jjˇ1 C

˛2jjˇ2 C � � � C

˛njjˇn :

Definition 6.30 The continued fraction (6.27) converges to C 2 C if the numbersCn are defined up to a finite subset of N0 and limn!1 Cn D C: In this case wewrite

C D ˇ0 C ˛1jjˇ1 C

˛2jjˇ2 C � � � C

˛njjˇn C : : : :

6.7 Continued Fractions 141

Proposition 6.31 For the numbers Cn we have Cn D AnBn

for n 2 N0, where

An D ˇnAn�1 C ˛nAn�2; A0 D ˇ0; A�1 D 1; (6.29)

Bn D ˇnBn�1 C ˛nBn�2; B0 D 1; B�1 D 0: (6.30)

If Bn ¤ 0 for n D 1; : : : ;m, then

Cm D Am

BmD ˇ0 C

mXnD1

.�1/nC1˛1 : : : ˛nBn�1Bn

: (6.31)

Proof The first assertion will be proved by induction on n. We define An and Bn

recursively by (6.29) and (6.30). Then we have to show that Cn D AnBn

.

For n D 0 this holds. Suppose that Cn D AnBn

. Recall that CnC1 is obtained if we

replace ˇn by ˇnC ˛nC1

ˇnC1in the formula for Cn. Then, using the induction hypothesis

and formulas (6.29) and (6.30), we derive

CnC1 D�ˇn C ˛nC1

ˇnC1

�An�1 C ˛nAn�2�

ˇn C ˛nC1

ˇnC1

�Bn�1 C ˛nBn�2

D ˇnC1.ˇnAn�1 C ˛nAn�2/C ˛nC1An�1ˇnC1.ˇnBn�1 C ˛nBn�2/C ˛nC1Bn�1

D ˇnC1An C ˛nC1An�1ˇnC1Bn C ˛nC1Bn�1

D AnC1BnC1

;

which completes the induction proof.Next we prove (6.31). Subtracting (6.29) multiplied by Bn�1 and (6.29) multi-

plied by An�1 we obtain

AnBn�1 � BnAn�1 D �˛n.An�1Bn�2 � Bn�1An�2/:

Repeated application of this relation yields

AnBn�1 � BnAn�1 D .�1/nC1˛1 : : : ˛n;

which can be written as

An

Bn� An�1

Bn�1D .�1/nC1˛1 : : : ˛n

Bn�1Bn: (6.32)

Now (6.31) follows by summing over n D 1; : : : ;m and using that A0B0D ˇ0. ut


Formulas (6.29) and (6.30) are three term recurrence relations that resemblethe relations for monic orthogonal polynomials. Equation (6.31) implies that thecontinued fraction converges if and only if the corresponding series in (6.31) does.

Now we will use the (positive definite) moment sequence s D .sn/n2N0 . If an; bndenote the corresponding Jacobi parameters an; bn, we set

ˇ0 D 0; ˇn D z � bn�1; ˛1 D s0; ˛nC1 D �a2n�1 for n 2 N:

Replacing n by nC 1 in (6.30) we obtain for the denominators

BnC1 D .z� bn/Bn � a2n�1Bn�1; B0 D 1;B�1 D 0; n 2 N;

and B1 D .z� b0/B0C s0 �B�1 D z� b0. These are precisely the recurrence relation(5.14) and the intial data for the monic orthogonal polynomials Pn.z/. Therefore,we have Bn D Pn.z/ for n 2 N0: In particular, B2 D .z � b1/.z � b0/� a20 D P2.z/.

For the numerators An we obtain in a similar manner

AnC1 D .z� bn/An � a2n�1An�1; A0 D 0;A�1 D 1; n 2 N;

and A1 D .z � b0/A0 C s0 � A�1 D s0: From Corollary 5.20 it follows thats�10 An satisfies the same recurrence relation and initial data as the corresponding

polynomials Qn.z/. Hence An D s0Qn.z/ for n 2 N0. In particular, A2 D.z � b1/s0 D s0Q2.z/.

Summarizing the preceding, we obtain

CnC1 D AnC1BnC1

D s0QnC1.z/PnC1.z/

D s0jjz � b0

C �a20jjz � b1

C � � � C �a2n�1j

jz� bn; n 2 N:

(6.33)

Now we assume that the moment sequence s is determinate. Let � be itsrepresenting measure. Then, by Markov’s theorem 6.29,

ZR

d�.x/

z � xD lim

n!1s0QnC1.z/PnC1.z/

; z 2 CnR:

Passing to the limit in (6.33) and inserting the latter equality yields

�I�.z/ DZR

d�.x/

z � xD s0jjz � b0

C �a20jjz � b1

C � � � C �a2n�1j

jz � bnC : : : ; z 2 CnR:

(6.34)

Formula (6.34) provides an expansion of the negative Stieltjes transform �I�.z/ of� as a continued fraction. It connects continued fractions and moment problems.We will not follow this path in this book and prefer to use other methods.

6.8 Exercises 143

Further, suppose that � is supported by a bounded interval Œa; b�. Then

�I�.z/ DZ

d�.x/

z � xD s0

zC s1

z2C � � � C sn

znC1 C � � � (6.35)

for jzj > max .jaj; jbj/. (Indeed, then 1z�x D

P1nD0 xn

znC1 converges uniformly onŒa; b�, so we can interchange summation and integration.) For general representingmeasures there is an asymptotic expansion of the form (6.35), see Proposition 7.12below.

The interplay between the two expansions (6.34) and (6.35) is one of theinteresting features of the classical theory of one-dimensional moment problems.

6.8 Exercises

1. Let s be a Hamburger moment sequence and � a representing measure for s.Prove that the following statements are equivalent:

(i) The moment problem for s is determinate.(ii) There exists a number z 2 CnR such that pz … l2.N0/ and qz … l2.N0/.

(iii) There is a number z0 2 CnR such that .x � z0/CŒx� is dense in L2.R; �/.(iv) There exist a number z0 2 CnR and a sequence .rn/n2N of polynomials

rn 2 CŒx� such that limn!1.x � z0/rn.x/ D 1 in L2.R; �/.

2. Let �; � 2MC.R/. Suppose that there exists a c > 0 such that �.M/ � c�.M/for each Borel subset M of R. Prove that if � is determinate, so is �.

3. Let s D .sn/n2N0 and t D .tn/n2N0 be moment sequences such that the momentsequence sC t WD .snCtn/n2N0 is determinate. Prove that s and t are determinate.

4. Let � 2MC.R/ and let a; b; c 2 R; a < c < b Suppose that supp� RnŒa; b�:Prove that � is determinate if and only if .x � c/CŒx� is dense in L2.R; �/:

5. Prove the “Plücker identity” for the Wronskian defined by (5.40):

W.˛; ˇ/nW.�; ı/n �W.˛; �/nW.ˇ; ı/n CW.˛; ı/nW.ˇ; �/n D 0; n 2 N0;

where ˛; ˇ; �; ı are arbitrary complex sequences.6. (Translation of Hamburger moment sequences)

Let s D .sn/n2N0 be a moment sequence and � 2 R. Define s.�/n DPnkD0

�nk

�� ksn�k for n 2 N0.

a. Show that s.�/ D .s.�/n/n2N0 is also a moment sequence.b. Show that � 2Ms if and only if �� 2Ms.�/, where d��.x/ WD d�.x � �/.c. Show that s is determinate if and only if s.�/ is determinate.

7. Let � 2 MC.R/ be symmetric. Assume that .�c; c/ \ supp� D ; for somec > 0. Show that the sequence

� qn.0/pn.0/

�n2N does not converge as n!1.


6.9 Notes

The Hamburger moment problem was first studied extensively by H. Hamburger[Hm]. Important classical result were obtained by M. Riesz [Rz2]. Among otherthings he discovered the operator-theoretic characterization of determinacy (Corol-lary 6.11). One-dimensional moment problems in the context of the extension theoryof symmetric operators were treated in [DM]. The terminology of a von Neumannsolution is from [Sim1]. Our approach partly follows [Sm9, Chapter 16] and [Sim1].

Markov’s theorem 6.29 goes back to A.A. Markov [Mv1]. It is proved in [Chi1]for measures with bounded support and in [VA] and [Be] for determinate measures.Our approach to Theorem 6.29 and Exercise 6.7 are taken from C. Berg [Be].

Continued fractions are treated in [Wl] and [JT].

Chapter 7The Indeterminate Hamburger MomentProblem

In this chapter we assume that s is an indeterminate Hamburger momentsequence. Our aim is to analyze the structure of the set Ms of all solutions ofthe moment problem for s. The central result in this respect is Nevanlinna’s theorem(Theorem 7.13) on the parametrization of Ms in terms of the set P[f1g, where Pare the holomorphic functions on the upper half plane with nonnegative imaginaryparts. It provides a one-to-one correspondence between Stieltjes transforms ofmeasures � 2 Ms and elements � 2 P [ f1g given by a fractional lineartransformation (7.16) with respect to four distinguished entire functions A;B;C;D.Note that in contrast to the moments the Stieltjes transform determines a finiteRadon measure uniquely (by Theorem A.13)!

In Sect. 7.1 these four Nevanlinna functions A;B;C;D are defined and investi-gated. Sections 7.2 and 7.5 contain fundamental results on von Neumann solutions(Theorems 7.6, 7.7, and 7.15). The family of Weyl circles is introduced in Sect. 7.3.In Sect. 7.4 the celebrated Nevanlinna theorem is proved. In Sect. 7.6 we givea short excursion into Nevanlinna–Pick interpolation and derive basic results onthe existence of a solution and on rational Nevanlinna functions (Theorems 7.20and 7.22). In Sect. 7.7 solutions of finite order are studied and a number ofcharacterizations of these solutions is given (Theorem 7.27 and 7.33).

7.1 The Nevanlinna Functions A.z/;B.z/;C.z/;D.z/

The crucial technical step for the definition of the Nevanlinna functions is containedin the following lemma.

Lemma 7.1 For any z 2 C the seriesP1

nD0 jpn.z/j2 andP1

nD0 jqn.z/j2 converge.The sums are uniformly bounded on compact subsets of the complex plane.

Proof Because the moment problem for s is indeterminate, it has at least twodifferent solutions � and �. The corresponding Stieltjes transforms I� and I�


145

146 7 The Indeterminate Hamburger Moment Problem

are holomorphic functions on CnR. Since � ¤ �, they do not coincide byTheorem A.13. Hence the set Z WDfz 2 CnR W I�.z/DI�.z/g has no accumulationpoint in CnR.

Let M be a compact subset of C. We choose b > 0 such that jzj � b for all z 2 Mand the line segment LWDfz 2 C W Im z D b; jRe zj � bg does not intersect Z . Bythe first condition and Corollary 5.31, the suprema of jpn.z/j and jqn.z/j over M areless than or equal to the corresponding suprema over L. Hence it suffices to provethe uniform boundedness of both sums on the set L.

Suppose that z 2 L. Then we derive

jI�.z/� I�.z/j2X

njpn.z/j2

DX

nj.qn.z/C I�.z/pn.z// � .qn.z/C I�.z/pn.z//j2

� 2X

njqn.z/C I�.z/pn.z/j2 C 2

Xnjqn.z/C I�.z/pn.z/j2

� 2b�1.Im I�.z/C Im I�.z// � 2b�1.jI�.z/j C jI�.z/j/;

where the inequality before last follows from inequality (5.39) in Proposition 5.37.Since the function jI�.z/�I�.z/j has a positive infimum on L (because L has apositive distance from the set Z) and I�.z/ and I�.z/ are bounded on L, the precedinginequality implies that the sum

Pn jpn.z/j2 is finite and uniformly bounded on L.

Using once more (5.39) and proceeding in a similar manner we derive

Xnjqn.z/j2 � 2

Xnjqn.z/C I�.z/pn.z/j2 C 2jI�.z/j2

Xnjpn.z/j2

� 2b�1jI�.z/j C 2jI�.z/j2X

njpn.z/j2

for z 2 L. Hence the boundedness of the sumP

n jpn.z/j2 on L implies theboundedness of

Pn jqn.z/j2 on L. ut

Lemma 7.2 For any sequence c D .cn/ 2 l2.N0/ the equations

f .z/ D1XnD0

cnpn.z/ and g.z/ D1XnD0

cnqn.z/ (7.1)

define entire functions f .z/ and g.z/ on the complex plane.

Proof We carry out the proof for f .z/. Since . pn.z// 2 l2.N0/ by Lemma 7.1 and.cn/ 2 l2.N0/, the series f .z/ converges for all z 2 C. We have

ˇˇ f .z/ �

kXnD0

cnpn.z/

ˇˇ2 D

ˇˇ 1XnDkC1

cnpn.z/

ˇˇ2 �

� 1XnDkC1

jcnj2�� 1X

nD0jpn.z/j2

�

7.1 The Nevanlinna Functions A.z/;B.z/;C.z/;D.z/ 147

for k 2 N and z 2 C. Therefore, since .cn/ 2 l2.N0/ and the sumP

n jpn.z/j2 isbounded on compact sets (by Lemma 7.1), it follows that

PknD0 cnpn.z/ ! f .z/

as k ! 1 uniformly on compact subsets of C. Hence f .z/ is holomorphic on thewhole complex plane. ut

From Lemmas 7.1 and 7.2 we conclude that

A.z;w/ WD .z�w/1XnD0

qn.z/qn.w/; B.z;w/ WD �1C .z�w/1XnD0

pn.z/qn.w/;

C.z;w/ WD 1C .z�w/1XnD0

qn.z/pn.w/; D.z;w/ WD .z�w/1XnD0

pn.z/pn.w/

are entire functions in each of the complex variables z and w. They are the limitsof the polynomials Ak.z;w/;Bk.z;w/;Ck.z;w/;Dk.z;w/, respectively, which havebeen defined in Proposition 5.24. By passing to the limit n!1 in formula (5.57)of Corollary 5.27 we obtain the important identity

A.z;w/D.z;w/ � B.z;w/C.z;w/ D 1: (7.2)

From the above formulas it is obvious that

A.z;w/ D �A.w; z/; B.z;w/ D �C.w; z/; D.z;w/ D �D.w; z/; z;w 2 C:

(7.3)

Definition 7.3 The four functions A.z;w/;B.z;w/;C.z;w/;D.z;w/ are called theNevanlinna functions associated with the indeterminate moment sequence s.

These four functions are a fundamental tool in the study of the indeterminatemoment problem. It should be emphasized that they depend only on the indetermi-nate moment sequence s.

We shall see by Theorems 7.6 and 7.13 below that the entire functions

A.z/ WD A.z; 0/;B.z/ WD B.z; 0/;C.z/ WD C.z; 0/;D.z/ WD D.z; 0/

will enter in the parametrization of solutions. Often these four entire functions A.z/,B.z/, C.z/, D.z/ are called the Nevanlinna functions associated with s.

A number of facts on these functions are collected in the next proposition. Scalarproducts and norms refer always to the Hilbert space l2.N0/.

Proposition 7.4 Suppose that z;w 2 C. Then we have:

(i) D.z; 0/B.w; 0/ � B.z; 0/D.w; 0/ D �D.z;w/.(ii) A.z;w/ D .z � w/hqz; qwi; D.z;w/ D .z � w/hpz; pwi.

B.z;w/C 1 D .z� w/hpz; qwi; C.z;w/ � 1 D .z � w/hqz; pwi:(iii) Im .B.z/D.z/ / D Im z kpzk2.(iv) D.z/ ¤ 0 and D.z/tC B.z/ ¤ 0 for z 2 CnR and t 2 R.(v) D.z/� C B.z/ ¤ 0 for all z; � 2 C, Im z > 0 and Im � � 0.


Proof

(i) follows from the formula (5.58) by passing to the limit n!1.(ii) Since pn and qn have real coefficients, pn.w/ D pn.w/ and qn.w/ D qn.w/.

Hence the formulas follow at once from the definitions of A;B;C;D and pz; qw.(iii) Using (i) and the second equality from (ii) we compute

B.z/D.z/ � B.z/D.z/ D D.z; z/ D .z�z/kpzk2:

(iv) Let z 2 CnR. Since p0.z/ ¤ 0, Im z kpzk2s ¤ 0 and hence D.z/ ¤ 0 by (iii).Assume to the contrary that D.z/tCB.z/ D 0 for some z 2 CnR and t 2 R.

Then we have �t D B.z/D.z/�1 and hence

0 D Im .B.z/D.z/�1jD.z/j2/ D Im .B.z/D.z// D Im z kpzk2

by (iii), which is a contradiction.(v) follows in a similar manner as the last assertion of (iv). ut

Proposition 7.5 If � 2Ms is a von Neumann solution, then

I�.z/ D � A.z;w/C I�.w/C.z;w/

B.z;w/C I�.w/D.z;w/for z;w 2 CnR: (7.4)

Formula (7.4) determines all values of I�.z/ on CnR provided one fixed valueI�.w/ is given.

Proof Since � is a von Neumann solution, Hs Š L2.R; �/. Hence fpn W n 2 N0g isan orthonormal basis of L2.R; �/, so by (5.38) for all z;w 2 CnR we have

fz D1XnD0

.qn.z/C I�.z/pn.z//pn; fw D1XnD0

.qn.w/C I�.w/pn.w//pn:

Using these formulas, the Parseval identity and Lemma 5.24 we derive

I�.z/ � I�.w/ D .z � w/Z

1

x � z

1

x � wd�.x/ D .z � w/ hfz; fwi�

D .z � w/1XnD0.qn.z/C I�.z/pn.z//.qn.w/C I�.w/pn.w/ /

D .z � w/1XnD0.qn.z/C I�.z/pn.z//.qn.w/C I�.w/pn.w//

D A.z;w/C I�.z/.B.z;w/C1/C I�.w/.C.z;w/�1/C I�.z/I�.w/D.z;w/:

Eliminating I�.z/ in the last equation we obtain (7.4). ut

7.2 Von Neumann Solutions 149

7.2 Von Neumann Solutions

Recall that the self-adjoint extensions of the Jacobi operator T on the Hilbert spaceHs are the operators Tt, t 2 R[f1g, from Theorem 6.23. If Et denotes the spectralmeasure of Tt, we set

�t.�/ WD s0 hEt.�/e0; e0i: (7.5)

Theorem 7.6 The measures�t , where t 2 R[f1g, are precisely the von Neumannsolutions for the indeterminate moment sequence s. For z 2 CnR, we have

s0h.Tt�zI/�1e0; e0i D I�t.z/ �ZR

1

x � zd�t.x/ D �A.z/C tC.z/

B.z/C tD.z/; (7.6)

where for t D1 the fraction on the right-hand side has to be set equal to �C.z/D.z/ .

Proof The measures �t are indeed the von Neumann solutions, since the operatorsTt exhaust the set of all self-adjoint extensions of T on Hs by Theorem 6.23.

The first equality of (7.6) follows at once from the definition of �t and thefunctional calculus for the resolvent of the self-adjoint operator Tt.

The main assertion of Theorem 7.6 is the last equality of (7.6). For this weapply formula (7.4) to � D �t, w D yi and pass to the limit y ! 0. Then theholomorphic function A.z;w/ in w tends to A.z; 0/ D A.z/. Similarly the limits ofB.z;w/;C.z;w/;D.z;w/ are B.z/;C.z/;D.z/, respectively.

Let t 2 R. Then limy!0 I�t.yi/ D t by Lemma 6.24 and D.z/tCB.z/ ¤ 0 onCnR by Proposition 7.4(iv), so the right-hand side of (7.4) tends to � A.z/CtC.z/

B.z/CtD.z/ :

Now let t D 1. Since limy!0 jI�1.yi/j D C1 by Lemma 6.24 and D.z/ ¤ 0

for z 2 CnR by Proposition 7.4(iv), in this case the limit of (7.4) is �C.z/D.z/ : ut

Since A.z/;B.z/;C.z/;D.z/ are entire functions, it follows from Eq. (7.6) that theStieltjes transform I�t.z/, t 2 R [ f1g, is a meromorphic function.

Further, the numerator and denominator in (7.6) have no common zero. Indeed,if t 2 R and z were a zero of AC tC and BC tD, then

A.z/D.z/ � B.z/C.z/ D .�tC.z//D.z/ � .�tD.z//C.z/ D 0;

which contradicts (7.2). Similarly, for t D1, C and D have no common zero.The following theorem describes the structure of von Neumann solutions.

Theorem 7.7 Suppose that s is an indeterminate Hamburger moment sequence.

(i) Each von Neumann solution �t of s has a discrete unbounded support. Thenumbers in supp�t are precisely the zeros of the entire function B.z/ C tD.z/for t 2 R resp. D.z/ for t D 1. The set supp�t is the spectrum of the self-adjoint operator Tt:


(ii) For each number x 2 R, there is a unique tx 2 R[f1g such that x 2 supp�tx .If t; Qt 2 R [ f1g and t ¤ Qt, then the supports of �t and �Qt are disjoint.

(iii) If x 2 supp�t, then x is a simple eigenvalue of the operator Tt and

�t.fxg/ D kpxk�2 D� 1X

nD0jpn.x/j2

��1: (7.7)

Proof

(i) By Proposition A.15, a closed subset K of R belongs to supp�t if and onlyif the Stieltjes transform I�t.z/ has a holomorphic extension to CnK. Hencesupp�t is the set of poles of the meromorphic function I�t . Since the numeratorand denominator in (7.6) have no common zero, these are precisely the zerosof the denominator. Being the zero set of an entire function, supp�t is discrete.

Since �t is a von Neumann solution, e0 is a cyclic vector for Tt. Hence Ttacts as the multiplication operator by the variable x in L2.R; �t/ and supp�t isthe spectrum of Tt; see [Sm9, Section 5.4]. The operator Tt is unbounded andso is its spectrum supp�t.

(ii) If D.x/ ¤ 0, then x is a pole of I�t for t D �B.x/D.x/�1, so that x 2 supp�t

by (i). Similarly, if D.x/ D 0, then x is a pole of I�1and hence x 2 supp�1

by (i).To prove the uniqueness assertion, assume that x 2 supp�t and x 2 supp�Qt

for t; Qt 2 R [ f1g. If t 2 R and Qt 2 R, then B.x/C tD.x/ D B.x/C QtD.x/ D0. Hence D.x/ ¤ 0 (otherwise AD � BC D 0; which contradicts (7.2)) andtherefore t D Qt. If t 2 R and Qt D 1, then B.x/C tD.x/ D 0 and D.x/ D 0, sothat B.x/ D 0. Again this contradicts (7.2). Thus, t D Qt.

(iii) Let x 2 supp�t: Then x is in the spectrum of Tt by (i). Because thisset is discrete, x is an eigenvalue of Tt. Since Tt T�, it follows fromProposition 6.6(i) that all corresponding eigenvectors are multiples of px D. pn.x/n2N0 . Thus kpxk�1px is a normalized simple eigenvector and the spectralprojection Et.fxg/ of Tt is the rank one projection Et.fxg/ D kpxk�2h�; pxi px.Since he0; pxi D p0.x/ D s�1=2

0 ; we obtain

�t.fxg/ D s0 hEt.fxg/e0; e0i D s0 kpxk�2he0; pxihpx; e0i D kpxk�2;

which proves (7.7). ut

Definition 7.8 A von Neumann solution of an indeterminate Hamburger momentsequence is called Nevanlinna extremal, or briefly, N-extremal.

By Theorem 7.7 (i) and (iii), each N-extremal solution � is the form

� D1XkD1

mkıxk ;

7.3 Weyl Circles 151

where mk > 0, the points xk 2 R are pairwise distinct, and limk!1 jxkj D 1.Further, fxk W k 2 Ng is the zero set of an entire function specified in Theorem 7.7(i).

7.3 Weyl Circles

An instructive tool in the theory of indeterminate moment problems is provided bythe Weyl circles.

Definition 7.9 For z 2 CnR the Weyl circle Kz is the closed circle in the complexplane with radius z and center Cz given by

z WD 1

jz�zj kpzk2sD z�zjz�zj

1

D.z; z/; Cz WD � .z�z/

�1 C hqz; pziskpzk2s

D �C.z; z/D.z; z/

:

The two equalities in this definition follow easily from Proposition 7.4(iii).The proof of the next proposition shows that in the indeterminate case the

inequality (5.39) means that the number I�.z/ belongs to the Weyl circle Kz.

Proposition 7.10 Suppose that � 2 Ms. Then the number I�.z/ lies in the Weylcircle Kz for each z 2 CnR. The measure � is a von Neumann solution if and onlyif I�.z/ belongs to the boundary @Kz for one (hence for all) z 2 CnR.

Proof We fix z 2 CnR and abbreviate � D I�.z/. The inequality (5.39) says that

kqzk2s C � hqz; pzis C � hqz; pzis C j�j2kpzk2s � kqz C �pzk2s �� z � z

: (7.8)

The inequality in (7.8) can be rewritten as

j�j2kpzk2s C ��hqz; pzisC.z�z/�1 �C � �hqz; pzisC.z�z/�1�C kqzk2s � 0 :

The latter inequality is equivalent to

kpzk2sˇ� C kpzk�2s .hqz; pzisC.z�z/�1/

ˇ2 � kpzk�2s jhqz; pzis C .z�z/�1j2 � kqzk2sand hence to

ˇ� C kpzk�2s .hqz; pzisC.z�z/�1/

ˇ2 � kpzk�4s � jhqz; pzis C .z�z/�1j2�kpzk2skqzk2s �:This shows that � lies in a circle with center Cz given above and radius

Qz D kpzk�2sqjhqz; pzis C .z�z/�1j2 � kpzk2skqzk2s (7.9)


provided the expression under the square root is nonnegative. Using Proposi-tion 7.4(iii), the relation C.z; z/ D �B.z; z/, and (7.2) we compute

Qz D z� z

D.z; z/

sjC.z; z/j2jz� zj2 �

A.z; z/D.z; z/

.z � z/2

D z� z

D.z; z/

s�C.z; z/B.z; z/jz� zj2 C A.z; z/D.z; z/

jz � zj2

D z�zD.z; z/

1

jz�zj D1

kpzk2s jz�zj:

Thus Qz is equal to the radius z of the Weyl circle and we have proved that � 2 Kz.The preceding proof shows that I�.z/ D � 2 @Kz if and only if we have equality

in the inequality (7.8) and hence in (5.39). The latter is equivalent to the relationfz 2 Hs by Proposition 5.21 and so to the fact that � is a von Neumann solution byProposition 6.13. This holds for fixed and hence for all z 2 CnR. ut

Let z;w 2 C. Since A.z;w/D.z;w/ � B.z;w/C.z;w/ D 1 by (7.2), the fractionallinear transformation Hz;w defined by

� D Hz;w.�/ WD �A.z;w/C �C.z;w/B.z;w/C �D.z;w/ (7.10)

is a bijection of the extended complex plane C D C[ f1g with inverse given by

� D H�1z;w.�/ D �

A.z;w/C �B.z;w/C.z;w/C �D.z;w/ : (7.11)

Some properties of these transformations Hz;w can be found in Exercises 7.5 and7.6. Here we will use only the transformations Hz WD Hz;0: Set R WD R [ f1g andrecall that CC D fz 2 C W Im z > 0g: The next lemma is illustrated in Fig. 7.1.

Fig. 7.1 The transformation Hz and the Weyl circle Kz

7.4 Nevanlinna Parametrization 153

Lemma 7.11

(i) Hz is a bijection of R onto the boundary @Kz D fI�t .z/ W t 2 Rg of the Weylcircle Kz for z 2 C nR.

(ii) Hz is a bijection of CC onto the interiorıKz of the Weyl circle Kz for z 2 CC.

(iii) Kz CC for z 2 CC.

Proof

(i) By Theorem 7.6, Hz maps R on the set fI�t .z/ W t 2 Rg. Since I�t.z/ 2 @Kz

by Proposition 7.10, Hz maps R into @Kz. But fractional linear transformationsmap generalized circles bijectively onto generalized circles. Hence Hz maps R

onto @Kz.(ii) From (i) it follows that Hz is a bijection of either the upper half-plane or the

lower half-plane on the interior of Kz. It therefore suffices to find one point

� 2 ıKz for which H�1

z .�/ 2 CC. Since I�0.z/; I�1.z/ 2 @Kz by (i),

� WD .I�0.z/C I�1.z//=2 D .�A.z/B.z/�1 � C.z/D.z/�1/=2 2 ı

Kz :

Here the second equality follows from Theorem 7.6. Inserting this expressioninto (7.11) we easily compute

H�1z .�/ D B.z/D.z/�1 D jD.z/j�2B.z/D.z/ :

Hence H�1z .�/ 2 CC by Proposition 7.4(iii), since z 2 CC.

(iii) Since z 2 CC, we have I�t .z/ 2 CC \ @Kz by (i). Hence Kz CC. ut

7.4 Nevanlinna Parametrization

First we prove a classical result due to Hamburger and Nevanlinna which is ofinterest in itself. It characterizes solutions of the moment problem in terms of theasympotic behaviour of their Stieltjes transforms.

Proposition 7.12

(i) If s D .sn/n2N0 is an arbitrary (!) Hamburger moment sequence and � 2Ms,then for each n 2 N0,

limy2R;y!1 ynC1

I�.iy/C

nXkD0

sk.iy/kC1

!D 0; (7.12)

where for fixed n the convergence is uniform on the set Ms.


(ii) Let s D .sn/n2N0 be a real sequence and I.z/ a Pick function. If (7.12) issatisfied (with I� replaced by I) for all n 2 N0, then s is Hamburger momentsequence and there exists a unique measure� 2Ms such that I.z/ D I�.z/ forz 2 CnR.

Proof

(i) First we rewrite the sum in Eq. (7.12) as

nXkD0

sk.iy/kC1

DnX

kD0

.�i/kC1ykC1

ZR

xkd�.x/D�i 1ynC1

ZR

nXkD0.�ix/kyn�k d�.x/

D �i 1ynC1

ZR

.�ix/nC1 � ynC1

�ix � yd�.x/ D 1

ynC1

ZR

.�ix/nC1 � ynC1

x � iyd�.x/:

(7.13)

Therefore, since jx � iyj�1 � jyj�1 for x; y 2 R, y ¤ 0, we obtain

ˇˇˇ ynC1

I�.iy/C

nXkD0

sk.iy/kC1

!ˇˇˇ

Dˇˇ Z

R

ynC1

x � iyd�.x/C

ZR

.�ix/nC1�ynC1

x � iyd�.x/

ˇˇ

DˇˇZR

.�ix/nC1

x � iyd�.x/

ˇˇ � jyj�1

ZR

jxjnC1 d�.x/ � jyj�1cn;

where cnWDsnC1 if n is odd and cnWDsn C snC2 if n is even. Since cn does notdepend on the measure �, we conclude that (7.12) holds uniformly on the setMs.

(ii) Condition (7.12) for n D 0 implies that limy!1 yI.iy/ D is0. Therefore,by Theorem A.14, the Pick function I is the Stieltjes transform I� of a finitepositive Radon measure � on R and � is uniquely determined. Since �.R/ <1, Lebesgue’s dominated convergence theorem applies and yields

s0 D limy!1�iyI.iy/ D lim

y!1�iyI�.iy/ D limy!1

ZR

�iyx � iy

d�.x/ DZR

d�.x/:

The main part of the proof is to show that the n-th moment of � exists andis equal to sn for all n 2 N0. We proceed by induction on n. For n D 0, this wasjust proved. Let n 2 N and assume that � has the moments s0; : : : ; s2n�2. Then,by the preceding proof of (i), formula (7.13) is valid with n replaced by 2n�2.We use this formula in the case 2n�2 to derive the second equality below and

7.4 Nevanlinna Parametrization 155

compute

.iy/2nC1� 2nX

kD0

sk.iy/kC1

C I�.iy/

�

D s2n C iys2n�1 C i2nC1y2�y2n�1

2n�2XkD0

sk.iy/kC1

C y2n�1I�.iy/�

D s2n C iys2n�1 C i2nC1y2�Z

R

.�ix/2n�1�y2n�1

x � iyd�.x/C

ZR

y2n�1

x � iyd�.x/

�

D s2n C iys2n�1 � y2ZR

x2n�1

x � iyd�.x/

D s2n �ZR

x2n

.x=y/2 C 1 d�.x/C iy

�s2n�1 �

ZR

x2n�1

.x=y/2 C 1 d�.x/

�:

By assumption (7.12) the term in the first line converges to zero as y!1.Considering the real part and using Lebesgue’s monotone convergence theoremwe get

s2n D limy!1

ZR

x2n

.x=y/2 C 1 d�.x/ DZR

x2nd�.x/ <1: (7.14)

The imaginary part, hence the imaginary part divided by y, also converges tozero as y!1. Since

jxj2n�1

.x=y/2 C 1 � jxj2n�1 � 1C x2n

and 1Cx2n is�-integrable by (7.14), the dominated convergence theorem yields

s2n�1 D limy!1

ZR

x2n�1

.x=y/2 C 1 d�.x/ DZR

x2n�1d�.x/: (7.15)

By (7.14) and (7.15) the induction proof is complete. utLet P denote the Pick functions (see Appendix A.2). We identify t 2 R with the

constant function t; then R becomes a subset of P. Set P WD P [ f1g.The main result in this section is the following theorem of R. Nevanlinna. It

expresses the Stieltjes transforms of representing measures of s by a fractionaltransformation of functions from the parameter space P.


Theorem 7.13 Suppose s is an indeterminate moment sequence. There is a one-to-one correspondence between functions � 2 P and measures � 2Ms given by

I�� .z/ �ZR

1

x � zd��.x/ D �A.z/C �.z/C.z/

B.z/C �.z/D.z/ � Hz.�.z//; z 2 CC: (7.16)

Proof Suppose that � 2Ms. If � is a von Neumann solution, then by Theorem 7.6there exists a t 2 R such that I�.z/ D Hz.t/ for all z 2 CC.

Assume that � is not a von Neumann solution and define �.z/ WD H�1z .I�.z//:

Let z 2 CC. Then I�.z/ 2ıKz by Proposition 7.10 and hence �.z/ D H�1

z .I�.z// 2CC by Lemma 7.11(ii). That is, �.CC/ CC. We show that C.z/CI�.z/D.z/ ¤ 0.Indeed, otherwise I�.z/ D �C.z/D.z/�1 D Hz.1/ 2 @Kz by Lemma 7.11(i)

which contradicts the fact that I�.z/ 2ıKz. Thus, �.z/ is the quotient of two

holomorphic functions on CC with nonvanishing denominator function. Therefore,� is holomorphic on CC. This proves that � 2 P. By the definition of � we haveHz.�.z// D I�.z/ on CC, that is, (7.16) holds.

Conversely, suppose that � 2 P. If � D t 2 R, then by Theorem 7.6 there is avon Neumann solution �t 2Ms such that I�t.z/ D Hz.t/.

Suppose now that � is not in R. Let z 2 CC and define I.z/ D Hz.�.z//. Then

�.z/ 2 CC and hence I.z/ D Hz.�.z// 2ıKz CC by Lemma 7.11 (ii) and (iii).

From Proposition 7.4(v) it follows that B.z/ C �.z/D.z/ ¤ 0. Therefore, I is aholomorphic function on CC with values in CC, that is, I 2 P.

To prove that I D I� for some � 2Ms we want to apply Proposition 7.12(ii).For this we have to check that condition (7.12) is fulfilled. Indeed, by Proposi-tion 7.12(i), given " > 0 there exists a Y" > 0 such that

ˇˇˇ ynC1

I�.iy/C

nXkD0

sk.iy/kC1

!ˇˇˇ < " for all y � Y" (7.17)

and for all � 2Ms. (Here it is crucial that Y" does not depend on � and that (7.17)is valid for all measures � 2Ms!) Fix a y � Y". Since I.iy/ D Hiy.�.iy// is in theinterior of the Weyl circle Kiy by Lemma 7.11(ii), I.iy/ is a convex combination oftwo points from the boundary @Kiy. By Lemma 7.11(i), all points of @Kiy are of theform I�t .iy/ for some t 2 R. Since (7.17) holds for all I�t .iy/ and I.iy/ is a convexcombination of values I�t .iy/, (7.17) remains valid if I�.iy/ is replaced by I.iy/. Thisshows that I.z/ fulfills the assumptions of Proposition 7.12(ii), so that I D I� forsome measure � 2Ms.

By Theorem A.13, the positive measure �� is uniquely determined by the valuesof its Stieltjes transform I�� on CC. Therefore, since I�� and � 2 P correspond toeach other uniquely by the relation I�.z/ D Hz.�.z// on CC, (7.16) gives a one-to-one correspondence between � 2Ms and � 2 P. ut

Let us briefly discuss and summarize some of the results obtained so far.

7.5 Maximal Point Masses 157

Theorem 7.13 provides a complete parametrization of the solution set Ms in theindeterminate case in terms of the set P. However, the one-to-one correspondence� $ �� between P and Ms given by (7.16) is highly nonlinear and very implicit.From Eq. (7.16) we derive that �.z/ is obtained from I�� .z/ by

�.z/ D � A.z/C I�� .z/B.z/

C.z/C I�� .z/D.z/; z 2 CC:

The subset R [ f1g of P is in one-to-one correspondence to the von Neumannsolutions, or equivalently, to the self-adjoint extensions of T on the Hilbert spaceHs Š l2.N0/. The nonconstant Pick functions correspond to self-adjoint extensionon a strictly larger Hilbert space (see Theorem 6.1).

Fix z 2 CC. The values I�.z/ for all von Neumann solutions � 2 Ms fill theboundary @Kz of the Weyl circle, while the numbers I�.z/ for all other solutions

� 2 Ms lie in the interiorıKz. By taking convex combinations of von Neumann

solutions it follows that each number ofıKz is of the form I�.z/ for some � 2Ms.

By Theorem 1.19, the solution set Ms is compact in the vague topology. Foreach indeterminate moment sequence the set Ms is “very large". We illustrate thisby stating two results without proofs from [BC1, Theorem 1]:

The subset of measures � 2Ms of the form d� D f .x/dx for some nonnegativefunction f 2 C1.R/ is dense in Ms with respect to the vague topology. The set ofmeasures of finite order (as defined in Sect. 7.7 below) is also dense inMs.

All solutions of finite order are extreme points of Ms (see Exercise 7.9). HenceMs is a convex compact set (in the vague topology) with dense set of extremepoints! Recall from Theorem 1.21 that a measure � 2 Ms is an extreme point ofMs if and only CŒx� is dense in L1.R; �/.

Remark 7.14 It is easily seen that the map 7! � WD � �1 is a bijection of theset P. Inserting this into (7.16) we obtain

I�� .z/ D �A.z/� .z/�1C.z/B.z/� .z/�1D.z/ D �

A.z/ .z/ � C.z/

B.z/ .z/ � D.z/; 2 P: (7.18)

The fraction on the right-hand side of (7.18) is another equivalent form ofparametrization of solutions which often occurs in the literature (for instance, inAkhiezer’s book [Ak]). Our convention (7.16) follows [Sim1]. ı

7.5 Maximal Point Masses

The following theorem and its subsequent corollary contain a remarkable propertyof von Neumann solutions concerning maximal point masses.


Theorem 7.15 Let � be a representing measure of the indeterminate momentsequence s. Suppose that � is not a von Neumann solution. Then, for any x 2 R

there is a von Neumann solution �t of s such that �t.fxg/ > �.fxg/:The following lemma is used in the proof of Theorem 7.15.

Lemma 7.16 Suppose that � 2 P and � … R. Let x 2 R and t 2 R. If the limitL.x; t/ WD lim"!C0 �.xCi"/�t

i"exists and is a real number, then L.x; t/ > 0.

Proof We use the canonical representation (A.4) of the Pick function

�.z/ D aC bzCZR

1C zt

t � zd�.t/;

where a; b 2 R, b � 0, and � is a finite positive Radon measure on R. Since L.x; t/is real by assumption, we derive

L.x; t/ D lim"!C0

�.xC i"/ � t

i"D lim

"!C0 Re�.xC i"/ � t

i"D lim

"!C0Im�.xC i"/

"

D lim"!C0

�bC

ZR

1C t2

.t � x/2 C "2 d�.t/�D bC

ZR

1C t2

.t � x/2d�.t/;

where the last equality holds by Lebesgue’s monotone convergence theorem. Theright-hand side is obviously non-negative. If it were zero, then we would have b D 0and � � 0, so that � 2 R, which contradicts the assumption. Thus L.x; t/ > 0. utProof of Theorem 7.15 First assume that D.x/ ¤ 0. Put t D �B.x/D.x/�1. Then wehave x 2 supp�t and �t.fxg/ > 0 by Theorem 7.7. Therefore the assertion is trivialif �.fxg/ D 0, so we can assume that �.fxg/ > 0.

Since A.x/D.x/� B.x/C.x/ D 1, we have

D.x/�1 D D.x/�1.A.x/D.x/� B.x/C.x// D A.x/C tC.x/: (7.19)

By Theorem 7.13 there is a unique � 2 P such that � D �� . Since � is not avon Neumann solution, � … R[ f1g. From the Stieltjes–Perron formula (A.8) andEq. (7.16) we obtain

�.fxg/ D lim"!C0 .�i"/I�.xC i"/ D lim

"!C0 i"A.xC i"/C �.xC i"/C.xC i"/

B.xC i"/C �.xC i"/D.xC i"/:

(7.20)

Therefore, since �.fxg/ > 0, we have lim"!C0 jI�.xC i"/j D C1 and hence

lim"!C0 �.xC i"/ D lim

"!C0 H�1xCi".I�.xC i"//

D lim"!C0 �

A.xC i"/C I�.xC i"/B.xC i"/

C.xC i"/C I�.xC i"/D.xC i"/D �B.x/

D.x/D t:

(7.21)

7.5 Maximal Point Masses 159

Now we compute the limits of the numerator and denominator of the fractionin the right-hand side of Eq. (7.20), where the factor i" is included into thedenominator. By (7.19) and (7.21) we obtain for the numerator

lim"!C0

�A.xC i"/C �.xC i"/C.xC i"/

� D A.x/C tC.x/ D D.x/�1: (7.22)

For the denominator we derive

lim"!C0

B.xC i"/C �.xC i"/D.xC i"/

i"

D lim"!C0

B.xC i"/ � B.x/C t.D.xC i"/ �D.x//CD.xC i"/.�.xC i"/ � t/

i"

D B0.x/C tD0.x/C D.x/ lim"!C0

�.xC i"/ � t

i": (7.23)

By (7.22), the limit of the numerator in the right-hand side of (7.20) exists and isreal. Therefore, since �.fxg/ > 0, the limit of the denominator exists as well. SinceD.x/ ¤ 0, it follows from (7.23) that the limit L.x; t/ WD lim"!C0 �.xCi"/�t

i"exists.

Since the limits in (7.20) and (7.22) are real, L.x; t/ is real. Hence L.x; t/ > 0 byLemma 7.16. Inserting the numerator and denominator limits into (7.20) we get

�.fxg/ D 1

D.x/.B0.x/C tD0.x//C D.x/2L.x; t/: (7.24)

On the other hand, we compute the mass �t.fxg/ by applying again formula(7.20) with � replaced by �t and � by t. Then, by (7.19), we obtain

�t.fxg/ D lim"!C0 .�i"/I�t.xC i"/ D lim

"!C0 i"A.xC i"/C tC.xC i"/

B.xC i"/C tD.xC i"/

D .A.x/C tC.x// lim"!C0

i"

B.xC i"/ � B.x/C t.D.xC i"/ � D.x//

D A.x/C tC.x/

B0.x/C tD0.x/D 1

D.x/.B0.x/C tD0.x//: (7.25)

Recall that D.x/ ¤ 0 and L.x; t/ > 0. Therefore, comparing (7.24) and (7.25) itfollows that �t.fxg/ > �.fxg/. This proves the assertion in the case when D.x/ ¤ 0.

If D.x/ D 0, then B.x/ ¤ 0 and the same proof goes through verbatim by usingthe second parametrization (7.18) of solutions. ut

The following corollary combines some assertions of Theorems 7.7 and 7.15.

Corollary 7.17 Let s be an indeterminate Hamburger moment sequence. For eachx 2 R there exists a unique von Neumann solution �t of s such that �t.fxg/ > 0.For any solution � ¤ �t of the moment problem for s we have �.fxg/ < �t.fxg/:


Proof The existence assertion on �t is contained in Theorem 7.7. Let � ¤ �t

be another solution. If � is a von Neumann solution, then �.fxg/ D 0 byTheorem 7.7(ii). If � is not a von Neumann solution, then �.fxg/ < �t.fxg/ byTheorem 7.15. utCorollary 7.18 Let � D P1

kD1mkıxk be an N-extremal solution of an indetermi-nate moment sequence, where all mk > 0 and the points xk 2 R are pairwisedistinct. Then, the measure �k WD ��mkıxk is determinate for k 2 N. In particular,� is the sum of the two determinate measures �k and mkıxk .

Proof Assume to the contrary that �k is indeterminate. By Corollary 7.17, there isa von Neumann solution �k of the moment sequence of �k such that �k.fxkg/ > 0.Clearly, � WD �k C mkıxk has the same moment sequence as � and �.fxkg/ > mk D�.fxkg/. This is impossible by the last assertion of Corollary 7.17. utRemark 7.19 Retain the assumptions of Corollary 7.18 and assume (upon trans-lation) that x1 D 0. Then sn.�/ D

Rxnd� D R

xnd�1 D sn.�1/ for n 2N. Thus, except for the first moment, the indeterminate measure � and thedeterminate measure�1 have the same moments and so the same growth of momentsequences! That is, there is no characterization of determinacy by growth conditionsof the moment sequence. The determinate moment sequence of �1 cannot satisfyCarleman’s condition (4.2), since otherwise � would be determinate by Carleman’stheorem 4.3. ı

For an indeterminate moment sequence s we define a function

s.z/ WD kpzk�2s D� 1X

nD0jpn.z/j2

��1; z 2 C:

This function plays an important role in the study of the moment problem for s.By Definition 7.9, z D jz � zj�1s.z/ is the radius of the Weyl circle Kz for

z 2 CnR. By Corollary 7.17 and Theorem 7.7(iii), for x 2 R the number s.x/is the maximal mass of the one point set fxg among all solutions of the momentproblem for s. This maximum is attained at a unique solution: the von Neumannsolution �t for which x 2 supp�t, that is, t D �B.x/D.x/�1 if D.x/ ¤ 0 and t D 1if D.x/ D 0.

7.6 Nevanlinna–Pick Interpolation

Recall that each Pick function ˚ 2 P has a representation

˚.z/ D aC bzCZR

�1

t � z� t

1C t2

�d�.t/; z 2 CnR; (7.26)

where a; b 2 R, b � 0, and � is a positive measure such thatR.1C t2/�1d�.t/ <1.

7.6 Nevanlinna–Pick Interpolation 161

Given a function W Z ! CC defined on a subset Z of the upper half-planeCC D fz 2 C W =z > 0g, the Nevanlinna–Pick interpolation problem asks:

When does there exist a function ˚ 2 P such that ˚.z/ D .z/ for all z 2 Z?

As in the case of moment problems, some appropriate positivity condition isnecessary and sufficient for the existence of a solution.

Theorem 7.20 Let Z be a subset of CC and W Z ! CC a function on Z . Thereexists a Pick function ˚ such that ˚.z/ D .z/ for all z 2 Z if and only if for eachfinite set Z of pairwise distinct elements z0; : : : ; zn 2 Z the matrix

K.Z/ WD�.zi/� .zj/

zi � zj

�n

i;jD0(7.27)

is positive semidefinite. If the set Z is finite, ˚ can be chosen rational.

Proof Assume first that there is a function ˚ 2 P such that ˚.z/ D .z/ for z 2 Z .Then using the canonical representation (7.26) we derive

.zi/� .zj/zi � zj

D ˚.zi/ �˚.zj/zi � zj

D bCZR

d�.x/

.x � zi/.x � zj/: (7.28)

Hence for �0; : : : ; �n 2 C we obtain

nXi;jD0


�i� j D b

ˇˇ nXiD0

�i

ˇˇ2 C

ZR

ˇˇ nXiD0

�i

x � zi

ˇˇ2d�.x/ � 0; (7.29)

which proves that the matrix K.Z/ is positive semidefinite.Now we prove the converse implication. Upon a linear transformation z 7! azCb

and a shift 7! � c with real a; b; c we can assume that i 2 Z and <.i/ D 0. Forz 2 CnR; let 'z denote the function

'z.x/ D 1C xz

x � z; x 2 R:

Next we verify that the functions f'z W z 2 CnR; z ¤ �ig are linearlyindependent overC. Indeed, suppose that �0'iCPk

jD1 �j'zj D 0, where z1; : : : ; zk 2CnR are distinct, zj ¤ �i; i, and �j 2 C. Note that 'i.x/ D i for x 2 R. Then weobtain

kXjD1

�j.1C z2j /.x � zj/�1 C

kXjD1

�jzj C �0i D 0; x 2 R:

Since zj ¤ �i; i and hence 1 � z2j ¤ 0 for j ¤ 0, the preceding equality implies that�j D 0 for j D 1; : : : ; k and hence also �0 D 0.


Since limx!˙1 'z.x/ D z, 'z is a continuous function on the one pointcompactification R D R [ f1g of R by setting '.1/ D z. Let E be the set offunctions

f .x/ D i�0'i CnX

jD1

��j'zj C �j'zj

�; (7.30)

where �0 2 R; �1; : : : ; �n 2 C and z1; : : : ; zn 2 Zni are distinct. Since 'z D 'z,each function f is real-valued and E is a real subspace C.RIR/. Since Z CC,it follows from the linear independence of the functions 'z shown above that thenumbers �j in (7.30) are uniquely determined by f . Hence there is a well-defined (!)linear functional L W E! R defined by

L. f / D �0i .i/CnX

jD1

��j.zj/C �j .zj/

�: (7.31)

Note that .i/ is defined, because i 2 Z , and i.i/ is real, since < .i/ D 0. Further,since .i/ 2 CC, the constant function g.x/ WD �i.i/ D =.i/ > 0 is in E.

Let LC W EC ! C be the extension of L to a linear functional on the complexvector space EC D EC iE. Clearly, 'z; 'z 2 EC for z 2 Z and (7.31) implies that

LC.'z/ D .z/ and LC.'z/ D .z/ for z 2 Z : (7.32)

The crucial step of this proof is to show that L. f / � 0 for f 2 EC. This is wherethe positivity assumption comes in. Suppose that the function f from (7.30) is inEC. Recall that 'i D i. Hence we can write f as

f .x/ D jx � z1j�2 : : : jx � znj�2p.x/

for some polynomial p. Since f .x/ � 0 on R, p.x/ � 0 on R. Therefore, byProposition 3.1, p D q21 C q22 for q1; q2 2 RŒx�. Setting q WD q1 C iq2 2 CŒx�,we have p.x/ D q.x/q.x/. Put

h.x/ WD .x � z1/�1 : : : .x � zn/

�1q.x/:

Since f is bounded on R, deg. p/ � 2n, so that deg.q/ � n. Therefore, since thenumbers zj are distinct, h.x/ is a linear combination of some constant and partialfractions .x � zj/�1. Setting z0 D i and using that .x � zj/�1 D .zj � i/�1. x�i

x�zj� 1/

for j D 1; : : : ; k, it follows that h.x/ can be written as

h.x/ DnX

jD0

x � i

x � zj�j (7.33)

7.6 Nevanlinna–Pick Interpolation 163

with �j 2 C. Then, inserting the corresponding functions, we compute

f .x/ D jh.x/j2 DnX

i;jD0

1C x2

.x � zi/.x � zj/�i� j D

nXi;jD0

'zi.x/� 'zj.x/zi � zj

�i � j: (7.34)

Therefore, by (7.32),

L. f / D LC. f / DnX

i;jD0


�i � j: (7.35)

Since the matrix K.Z/ is positive semidefinite by (7.27), L. f / � 0 by (7.35).Thus L and E satisfy the assumptions of Proposition 1.9, so there exists a (finite)

measure Q� 2 MC.R/ such that L. f / D Rfd Q� for f 2 E. Define �.M/ D Q�.M/ for

a Borel subset M of R and b D Q�.f1g/. Then � is a finite measure of MC.R/ and

L. f / D bf .1/CZR

f .x/ d�.x/; f 2 E: (7.36)

Clearly, (7.36) extends to LC. f / and f 2 EC. By (7.32), LC.'z/ D .z/ and hence

.z/ D LC.'z/ D bzCZR

1C xz

x � zd�.x/; for z 2 Z : (7.37)

The right-hand side of (7.37) defines a function ˚ 2 P. This completes the proofof the converse implication.

If the set Z is finite, the vector space E is finite-dimensional. Then Proposi-tion 1.26 applies instead of Proposition 1.9 and yields a finitely atomic measure Q�.Then (7.37) gives a rational function ˚ 2 P. utCorollary 7.21 Let z0; : : : ; zn 2 CC be pairwise distinct and w0; : : : ;wn 2 CC.There exists a function ˚ 2 P such that ˚.zj/ D wj for j D 0; : : : ; n if and only ifthe matrix

K D .Kij/nijD0; where Kij D wi � wj

zi � zj; (7.38)

is positive semidefinite. In this case the function ˚ can be chosen rational.

Proof We apply Theorem 7.20 with Z WD fz0; : : : ; zng and .zj/ WD wj, j D0; : : : ; n. Then the necessity of the positive semidefiniteness of K is stated inTheorem 7.20 and its sufficiency follows from the preceding proof. Indeed, by thedefinition of Z all functions f of E are of the form (7.30), so by the above proof itsuffices that the single matrix K is positive semidefinite. ut


The degree of a rational function '.z/ D p.z/q.z/ , where p; q 2 CŒz� have no common

zero, is defined by

deg.'/ WD max.deg. p/; deg.q//: (7.39)

Theorem 7.22 Let z0; : : : ; zn 2 CC be pairwise distinct and w0; : : : ;wn 2 CC.Suppose that there exists a ˚ 2 P such that ˚.zj/ D wj for j D 0; : : : ; n. Then thefollowing are equivalent:

(i) The matrix (7.38) has the eigenvalue 0.(ii) ˚ is real rational function with degree at most n.

(iii) The interpolating function ˚ 2 P is uniquely determined.

Proof For �0; : : : ; �n 2 C we compute (see (7.29))

nXi;jD0

Kij �i� j DnX

i;jD0

˚.zi/ �˚.zj/zi � zj

�i� j D b

ˇˇ nXiD0

�i

ˇˇ2 C

ZR

ˇˇ nXiD0

�i

x � zi

ˇˇ2d�.x/:

(7.40)

(i)!(ii) Let .�0; : : : ; �n/ be an eigenvector of K for the eigenvalue 0. Then theexpression in (7.40) is zero. Set p.x/ D Qn

iD0.x � zi/. Then there is a polynomial

q of degree at most n such thatPn

iD0�j

x�ziD q.x/

p.x/ : Since (7.40) vanishes, we

obtainR ˇ q.x/

p.x/

ˇ2d�.x/ D 0. Therefore, � is supported on the zero set Z.q/ by

Proposition 1.23. This has at most n points. If b ¤ 0, we have in additionPniD0 �i D 0 by (7.40). This implies that deg.q/ � n � 1, so that � is supported

at at most n� 1 points. In both cases b D 0 and b ¤ 0 it follows from the canonicalrepresentation (7.26) that ˚ is a real rational function of degree at most n.

(ii)!(i) Suppose that ˚ is rational and deg.˚/ � n. Since ˚ is holomorphicoutside the support of �, it follows at once from (7.26) that � is supported at k pointsx1; : : : ; xk, where k � n if b D 0 and k � n � 1 if b ¤ 0. Put m D k if b D 0 andm D k C 1 if b ¤ 0. In either case m � n. For � D .�0; : : : ; �n/ 2 CnC1 we defineh.�/ D .h1.�/; : : : ; hm.�// 2 Cm, where

hj.�/ DnX

iD0

�i

xj � zi; j D 1; : : : ; k;

and hm.�/ D PniD0 �i if b ¤ 0. Since m � n, the mapping h W CnC1 ! Cm has a

nontrivial kernel. If � ¤ 0 is in this kernel, thenPn

i;jD0 Kij�i� j D 0 by (7.40). Sincethe matrix (7.38) is positive semidefinite by Corollary 7.21, we conclude that � is aneigenvector for the eigenvalue 0.

(i)!(iii) Let Q be a solution of the interpolation problem. Then Q is a realrational function of degree at most n by (i)!(ii). Since Q is real on R, we haveQ .z/ D Q .z/ for z 2 CnR by Schwarz’ reflection principle. Hence Q .zj/ D wj and

7.7 Solutions of Finite Order 165

Q .zj/ D wj for j D 0; : : : ; n: Hence Q is uniquely determined, because a rationalfunction of degree at most n is determined by 2n C 1 distinct points. (Indeed, ifq1p1.vj/ D q2

p2.vj/ for distinct v1 : : : ; v2nC1, then q1.vj/p2.vj/�q2.vj/p1.vj/ D 0. Since

deg.q1p2 � q2p1/ � 2n, this implies q1p2 � q2p1 D 0, so that q1p1D q2

p2.)

(iii)!(i) Assume to the contrary that 0 is not an eigenvalue of the matrix (7.38).We use the setup of the proof of Theorem 7.20. Let E C.RIR/ be the real vectorspace of functions (7.30) with n fixed. Suppose that f 2 EC and L. f / D 0. Since0 is not an eigenvalue, the matrix (7.38) is positive definite. Therefore, by (7.35),L. f / D 0 implies that all numbers �j in (7.33) are zero, so that h D 0 and hencef D 0. Thus, L is strictly EC-positive. Hence, by Theorem 1.30(ii), L has differentrepresenting measures Q� and these measures give different interpolating functions˚by the right-hand side of (7.37). This contradicts (iii) and completes the proof. ut

There is an alternative proof of the last implication (iii)!(i). One can replaceTheorem 1.30(ii) by Theorem 9.7 (proved in Sect. 9.1 below) on the truncatedmoment problem. Let us sketch the necessary modifications. First we note thatdimE D 2nC1. (To see this it suffices to recall that the complex space EC is spannedby the linearly independent functions 1; 'zj ; 'zj , j D 1; : : : ; n, so it has dimension2nC1.) For each f 2 E there exists a unique polynomial pf 2 RŒx�2n such that

f .x/ D jx � z1j�2 : : : jx � znj�2pf .x/:

Since dimRŒx�2n D dimE D 2nC1, the map f 7! pf is a linear bijection of E ontoRŒx�2n: Hence there exists a linear functional on RŒx�2n defined by QL. pf / D L. f /,f 2 E: Let q 2 RŒx�n, q ¤ 0. Then q2 D pf for some f 2 E. Since q ¤ 0, we havef ¤ 0 and f 2 EC, so that L. f / > 0 as shown in the preceding proof of (iii)!(i).Thus, QL.q2/ D QL. pf / D L. f / > 0. Therefore, by Theorem 9.7, there exists a one-parameter family �t of finitely atomic representing measures for the functional QL.Hence, setting d�t D jx� z1j2 : : : jx� znj2d�t, we obtain a one-parameter family ofrepresenting measures for L and so of interpolating functions˚ by (7.37).

7.7 Solutions of Finite Order

Throughout this section we assume that � is a measure in MC.R/. This meansthat all moments of � are finite. Let s denote its moment sequence. Recall fromProposition 6.2 that the canonical Hilbert space Hs is a closed subspace of L2.R; �/.

Definition 7.23 For a measure � 2MC.R/ the order of � is defined by

ord.�/ D dim .L2.R; �/ Hs/: (7.41)

Here dim means the cardinality of an orthonormal basis of the Hilbert space.


Comparing Definitions 6.4 and 7.23 we see that the von Neumann solutions areprecisely the solutions of order 0. Thus, � 2 MC.R/ has order 0 if and only ifeither its moment sequence is determinate (by Theorem 6.10) or it is an N-extremalsolution (Definition 7.8) of an indeterminate moment sequence. Therefore, for eachmeasure of nonzero order the corresponding moment sequence is indeterminate.

Measures of finite order are, after N-extremal measures, the simplest solutions ofindeterminate moment problems. In this section we derive a number of characteri-zations of such measures.

We begin with some preliminaries. A crucial role is played by the functions

fz.x/ D 1

x � z; where z 2 CnR; x 2 R:

For n1; : : : ; nk 2 N0 and pairwise distinct z1; : : : ; zk 2 CnR we define a closedlinear subspace of L2.R; �/ by

H.z1; : : : ; zkI n1; : : : ; nk/ WD Hs C Linf f jzl W j D 1; : : : ; nl; l D 1; : : : ; kg: (7.42)

(If nl D 0 for some l, the corresponding term in (7.42) will be set zero.) Further, weshall use the bounded operator V.z;w/ of L2.R; �/ with bounded inverse defined by

V.z;w/ WD .A� � zI/.A� � wI/�1 D I C .w � z/.A� � wI/�1; z;w 2 CnR;

where A� is the multiplication operator by the variable x on L2.R; �/, see (6.2).

Lemma 7.24

(i) .A� � zI/�1Hs Hs C C � fz for z 2 CnR:(ii) f kz f

mw 2 Lin f f jz ; f lw W jD1; : : : ; k; lD1; : : : ;mg for z;w 2 CnR; z ¤ w; k; n 2 N:

(iii) Suppose that z1; : : : ; zk 2 CnR are pairwise distinct and n1; : : : ; nk 2 N0. IffnjC1zj 2H.z1; : : : ; zkI n1; : : : ; nk/; then f kzj2H.z1; : : : ; zkI n1; : : : ; nk/ for k 2 N:

Proof

(i) Let p 2 CŒx�. Then qz.x/ WD p.x/�p.z/x�z is a polynomial in x and

..A� � zI/�1p/.x/ D .x � z/�1p.x/ D p.x/� p.z/

x � zC p.z/.x � z/�1

D qz.x/C p.z/fz.x/ 2 Hs C C � fz:

Thus .A��zI/�1CŒx� HsCC � fz. Since HsCC � fz is closed in L2.R; �/, CŒx�is dense in Hs and .A� � zI/�1 is bounded, it follows that .A� � zI/�1Hs Hs C C � fz.

(ii) The assertion follows by induction on jC l easily from the identity

f jz flw D . f jz f l�1w � f j�1z f lw/.z� w/�1; j; l 2 N:


(iii) For notational simplicity let j D 1 and write z D z1; n D n1.First let n � 1. By the assumption we can write f nC1

z D PnlD1 �lf lz C g,

where �l 2 C and g 2 H.z2; : : : ; zkI n2; : : : ; nk/. Applying .A� � zI/�1 weobtain

f nC2z D

nXlD1

�lflC1z C .A� � zI/�1g: (7.43)

From (i) and (ii) it follows that .A� � zI/�1g 2 H.z; z2; : : : ; zkI 1; n2; : : : ; nk/:Since f nC1

z 2 H.z; : : : ; zkI n; : : : ; nk/ by assumption,Pn

lD1 �lf lC1z 2H.z; : : : ; zkI n; : : : ; nk/: Thus, if n � 1, both summands in (7.43) are inH.z; : : : ; zkI n; : : : ; nk/ and so is f nC2

z .Let n D 0. Then fz 2 H.z2; : : : ; zkI n2; : : : ; nk/ by assumption, so by (i)

and (ii),

f 2z 2 C � fz CH.z2; : : : ; zkI n2; : : : ; nk/ H.z; z2; : : : ; zkI 0; n2; : : : ; nk/:

This completes the proof of the assertion for k D n C 1. Proceeding in asimilar manner by induction it follows that f kz 2 H.z; : : : ; zkI n; : : : ; nk/ for allk 2 N: ut

Lemma 7.25 Let w; z1; : : : ; zk 2 CnR be pairwise distinct and n1; : : : ; nk 2 N0.Then, for l D 1; : : : ; k, we have

.i/ V.zl;w/H.z1; : : : ; zkI n1; : : : ; nl C 1; : : : ; nk/ D H.z1; : : : ; zk;wI n1; : : : ; nk; 1/:.ii/ V.w; zl/H.z1; : : : ; zk;wI n1; : : : ; nk; 1/ D H.z1; : : : ; zkI n1; : : : ; nl C 1; : : : ; nk/:

Proof From Lemma 7.24(ii) we conclude that V.w; zl/ and V.zl;w/ map thecorresponding subspaces into the spaces on the right. Therefore, the compositionI D V.zl;w/V.w; zl/ maps H.z1; : : : ; zk;wI n1; : : : ; nk; 1/ into, hence onto, itself.This implies that we have equality in (ii) and similarly in (i). utProposition 7.26 Suppose � 2 MC.R/ and ord.�/ D n 2 N. Let z1; : : : ; zk 2CnR be pairwise distinct and n1; : : : ; nk 2 N such that n D n1 C � � � C nk. Then

L2.R; �/ D Hs C Lin f f jzl W j D 1; : : : ; nl; l D 1; : : : ; kg:

Proof Let A D Lin f f kz W z 2 CnR; k 2 Ng. By Lemma 7.24(ii), A is closedunder multiplication, so A is a �-subalgebra of the C�-algebra C0.R/ of continuousfunctions on R vanishing at infinity. Obviously, A separates the points of R.Hence, by the Stone–Weierstrass theorem [Cw, Corollary 8.3], A is norm densein C0.R/. Since the measure � is finite, this implies that A is dense in L2.R; �/.Since ord.�/ D n, there exists a finite-dimensional subspace B of A such thatL2.R; �/ D Hs C B. The latter implies that L2.R; �/ D Hs CA.


Let fv1; : : : ; vrg be a maximal subset of CnR such that F1 WD f fv1 ; : : : ; fvrg islinearly independent modulo Hs, where maximality means that fv 2 Hs C LinF1for any v 2 CnR distinct from all vj. It is clear that such a set always exists, sinceord.�/ D n and L2.R; �/ D Hs CA. Further, there exist numbers m1; : : : ;mr 2 N

such that F D f f jvl W j D 1; : : : ;ml; l D 1; : : : ; rg is a maximal set which is linearlyindependent modulo Hs. Here the maximality means that f mlC1

vl2 Hs C LinF for

all l D 1; : : : ; r. Then, it follows from the definition (7.42) that

G WD H.v1; : : : ; vrIm1; : : : ;mr/ D Hs C LinF:

If z 2 CnR, z ¤ vj for all j, then fz 2 G and hence f kz 2 G for all k 2 N

by Lemma 7.24(iii). Likewise, fmjC1vj 2 G implies f kvj 2 G for k 2 N, again by

Lemma 7.24(iii). This proves that A G. Since L2.R; �/ D Hs C A as shown inthe preceding paragraph, we get L2.R; �/ D G. Because F is linearly independentmodulo Hs,

n D ord.�/ D dimL2.R; �/=Hs D dim LinF D m1 C � � � C mr:

Thus, by the preceding we have shown that

L2.R; �/ D H.v1; : : : ; vrIm1; : : : ;mr/ D Hs C Lin ff jvl W j D 1; : : : ;ml; l D 1; : : : ; rg:This equality is of the required form except for the fact that we have to take ourgiven functions fzl and numbers nl instead of fvl and ml, respectively. To remedy thiswe now use Lemma 7.25.

First let us choose w1; : : : ;wn 2 CnR such that both w1; : : : ;wn; v1 : : : ; vr andw1; : : : ;wn; z1; : : : ; zk are pairwise distinct. Then, by Lemma 7.25(i), we can finda products of operators V.vl;wi/ that maps L2.R; �/ D H.v1; : : : ; vrIm1; : : : ;mr/

ontoH.w1; : : : ;wnI 1; : : : ; 1/. Further, by Lemma 7.25(ii), there is a product of oper-ators V.wi; zj/ which maps H.w1; : : : ;wnI 1; : : : ; 1/ onto H.z1; : : : ; zkI n1; : : : ; nk/.Since all operators V.vl;wi/ and V.wi; zj/ are isomorphisms of L2.R; �/, we obtain

L2.R; �/ D H.z1; : : : ; zkI n1; : : : ; nk/ D Hs C Linf f jzl W jD1; : : : ; nl; lD1; : : : ; kg: ut

Proposition 7.26 says that f f jzl W j D 1; : : : ; nl; l D 1; : : : ; kg forms abasis of the quotient space L2.R; �/=Hs if ord.�/ D n1C : : :Cnk 2 N andz1; : : : ; zk 2 CnR are pairwise distinct. This is a crucial step for the followingtheorem, which characterizes measures of finite order in terms of density anddeterminacy conditions.

Theorem 7.27 Suppose that � 2 MC.R/. Let z1; : : : ; zk 2 CnR be pairwisedistinct numbers and z 2 CnR. Let n1; : : : ; nk 2 N and set n D n1 C � � � C nk.Define measures �n and �nC1 of MC.R/ by

d�n.x/ DkY

lD1jx � zlj�2nld�.x/; d�nC1.x/ D jx � zj�2d�n.x/:


Then the following statements are equivalent:

(i) ord.�/ � n.(ii) L2.R; �/ D Hs C Linf f jzl W j D 1; : : : ; nl; l D 1; : : : ; kg:

(iii) CŒx� is dense in L2.R; �n/.(iv) �nC1 is determinate.

Proof In this proof we abbreviate F D Linf f jzl W j D 1; : : : ; nl; l D 1; : : : ; kg:(i)!(ii) Assume that m WD ord.�/ � n. If m D 0, then L2.R; �/ D Hs and

the assertion trivially holds. Now let m 2 N. We choose natural numbers i � k andmj � nj for j D 1; : : : ; i such that m D m1 C � � � C mi. Then, by Proposition 7.26,L2.R; �/ D Hs C Lin f f jzl W j D 1; : : : ;ml; l D 1; : : : ; ig: By adding further powersf jzl if m < n this obviously implies L2.R; �/ D Hs C F ; which proves (ii).

(ii)!(i) is trivial, since dim F � n1 C � � � C nk D n:(ii)$(iii) From the definition of the measure �n we see that the mapping U

defined by .Uf /.x/ D QklD1.x � zl/nl f .x/ is a unitary operator of L2.R; �/ onto

L2.R; �n/. Decomposition into partial fractions yields an identity

kYlD1.x � zl/

�nl DkX

lD1

nlXjD1

alj.x � zl/j

; (7.44)

where alj 2 C. Equation (7.44) implies that U maps CŒx� C F into CŒx�. FromLemma 7.24(ii) it follows that U maps CŒx� C F onto CŒx�. Since CŒx� is densein Hs, (ii) is equivalent to the density of CŒx� C F in L2.R; �/: Hence CŒx� C Fis dense in L2.R; �/ if and only if U.CŒx� C F/ D CŒx� is in L2.R; �n/. Thus,(ii)$(iii).

(iii)$(iv) By Theorem 6.13, CŒx� is dense in L2.R; �n/ if and only if thefunction fz D .x�z/�1 is in the closure of CŒx� in L2.R; �n/. From Corollary 6.12(or from Exercise 6.1) it follows that �nC1 is determinate if and only if 1 is in theclosure of .x � z/CŒx� in L2.R; �nC1/. But both conditions are equivalent, since

Zj.x � z/�1 � p.x/j2d�n.x/ D

Zj1 � .x � z/p.x/j2jx � zj�2d�n.x/

DZj1 � .x � z/p.x/j2d�nC1.x/ for p 2 CŒx�: ut

Since statement (i) of Theorem 7.27 does not depend on the choice of zj; nj, thisholds for the assertions (ii)–(iv) as well. We elaborate on this in some corollaries.

Corollary 7.28 Let � 2MC.R/, wj 2 CnR for j 2 N: Define �k 2MC.R/ by

d�k WDkY

jD1jx � wjj�2d�; k 2 N; �0 WD �: (7.45)


Then the following statements are equivalent:

(i) ord.�/ is finite.(ii) There exists a k 2 N such thatCŒx� is dense in L2.R; �k/ for some, equivalently

for arbitrary, numbers w1; : : : ;wk 2 CnR.(iii) There exists an m 2 N such that �m is determinate for some, equivalently for

arbitrary, numbers w1; : : : ;wm 2 CnR.

Further, ord.�/ � k if (ii) holds and ord.�/ � m � 1 if (iii) is satisfied.

Proof Let w1; : : : ;wn 2 CnR be arbitrary. We denote by z1; : : : ; zk the pairwisedistinct ones among them and by nj the multiplicity of wj in the sequencefw1; : : : ;wng. Then we are in the setup of Theorem 7.27 and the equivalence ofstatements (i), (iii), (iv) therein yields the assertions. ut

The next corollary follows at once from the last statement in Corollary 7.28.

Corollary 7.29 Retain the assumptions and the notation of Corollary 7.28, that is,wj 2 CnR are arbitrary, and �k is defined by (7.45). Let n 2 N. Then:

(i) ord.�/ D n if and only if CŒx� is dense in L2.�n/, but not in L2.�n�1/.(ii) ord.�/ D n if and only if �nC1 is determinate, but �n is not determinate.

Corollary 7.30 For a measure � 2MC.R/ the following are equivalent:

(i) ord.�/ is finite.(ii) L2.R; �/ D Hs C Linf fw1 ; : : : ; fwng for some n 2 N and some, equivalently

for arbitrary, pairwise distinct numbers w1; : : : ;wn 2 CnR.(iii) L2.R; �/ D Hs C Linf fw; f 2w; : : : ; f nwg for some n 2 N and some, equivalently

for arbitrary, w 2 CnR.

Proof We regroup fw1; : : : ;wng as in the proof of Corollary 7.28 and applyTheorem 7.27 (i)$(ii). utCorollary 7.31 If� 2MC.R/ has order n 2 N, then the support of� is a discreteunbounded set.

Proof We retain the notation (7.45). By Corollary 7.29 (i) and (ii), CŒx� is densein L2.R; �n/ and �n is not determinate. Thus �n is a von Neumann solution of anindeterminate moment problem. Hence the support of �n is discrete and unboundedby Theorem 7.7. Since d� DQn

jD1 jx � wjj2d�n.x/, so is the support of �. utThe next proposition relates measures of finite order to moment problems with

constraints on their Stieltjes transforms.Let z1; : : : ; zn 2 CC be pairwise distinct and w1; : : : ;wn 2 CC. We abbreviate

r.x/ D jx � z1j2 � � � jx � znj2:

For p 2 CŒx� there is a decomposition of the rational function pr as a sum of partial

fractions


p.x/

r.x/D qp.x/C

nXjD1

�aj;px � zj

C bj;px � zj

�; (7.46)

where aj;p; bj;p 2 C and qp 2 CŒx�. The constants aj;p; bj;p and the polynomial qp areuniquely determined by p and z1 : : : ; zn. Note that aj;p D bj;p and qp D qp.

For a linear functional L on CŒx�, we define a linear functional on CŒx� by

�.L/.z1;:::;znIw1;:::;wn/. p/ D L.qp/CnX

jD1.aj;pwj C bj;pwj/; k 2 N0: (7.47)

Because of (7.46) the functional L and the numbers wj can be recovered from thefunctional�.L/.z1;:::;znIw1;:::;wn/ by the formulas

L. p.x// D �.L/.z1;:::;znIw1;:::;wn/. p.x/r.x//; (7.48)

wj D �.L/.z1;:::;znIw1;:::;wn/.r.x/.x � zj/�1/: (7.49)

(The latter expression is well-defined, since r.x/.x � zj/�1 is a polynomial.)

Proposition 7.32 Suppose that s is a moment sequence. Let z1; : : : ; zn 2 CC bepairwise distinct and w1; : : : ;wn 2 CC: There is a bijection between all solutions� 2 Ms satisfying I�.zj/ D wj for j D 1; : : : ; n and solutions �n 2 MQs, whereQs D .Qsk/k2N0 and Qsk WD �.Ls/.z1;:::;znIw1;:::;wn/.x

k/, k 2 N0, is defined by (7.47) forp D xk. (Note that both sets of solutions may be empty.) This bijection is given by

d� $ d�n WDnY

jD1jx � zjj�2d� � r.x/�1d�: (7.50)

Proof First, let � 2 Ms be such that I�.zj/ D wj for j D 1; : : : ; n. Let k 2 N0.Then, using (7.50), (7.46), and finally (7.47), we derive

Zxkd�n.x/ D

Zxkr.x/�1d�.x/

DZ

qxk.x/d�.x/CnX

jD1

�aj;xk

Zd�.x/

x � zjC bj;xk

Zd�.x/

x � zj

�

D �.Ls/.z1;:::;znII�.z1/;:::;I�.zn//.xk/ D �.Ls/.z1;:::;znIw1;:::;wn/.xk/ D Qsk;

that is, �n 2MQs.Now suppose that �n 2MQs. Using the formulas (7.48) and (7.49) we obtain

sk D Ls.xk/ D �.Ls/.z1;:::;znIw1;:::;wn/.x

kr.x// DZ

xkr.x/d�n.x/ DZ

xkd�.x/;


wj D �.Ls/.z1;:::;znIw1;:::;wn/

�r.x/.x � zj/

�1/

DZ

r.x/.x � zj/�1 d�n.x/ D

Z.x � zj/

�1d�.x/ D I�.zj/

for k 2 N0 and j D 1; : : : ; n. Thus, � 2Ms and I�.zj/ D wj; j D 1; : : : ; n: utThe next theorem states that solutions of finite order are precisely those

corresponding to rational functions ˚ 2 P in the Nevanlinna parametrization ofTheorem 7.13. In the proof we use Theorem 7.22 on Nevanlinna–Pick interpolation.

Theorem 7.33 Suppose that s is an indeterminate Hamburger moment sequence. Arepresenting measure �˚ 2Ms (given by (7.16) with ˚ 2 P) has a finite order ifand only if ˚ is a rational function. In this case, ord.�˚/ D deg.˚/; where deg.˚/is defined by (7.39).

Proof We fix z1; : : : ; znC1 2 CC pairwise distinct and abbreviate � WD �˚ .As in Theorem 7.27 and Proposition 7.32 we define �nC1 2MC.R/ by

d�nC1.x/ DnC1YjD1jx � zjj�2d�.x/:

Then vj WD I�.zj/ and wj WD H�1zj .vj/ are in CC for j D 1; : : : ; nC 1. By (7.16) we

have Hzj.˚.zj// D I�.zj/ D vj and hence ˚.zj/ D H�1zj .vj/ D wj: That is, ˚ is a

solution of the Nevanlinna–Pick interpolation problem

�.zj/ D wj; j D 1; : : : ; nC 1; for � 2 P: (7.51)

Suppose � has order n 2 N. Let Q 2 P be another solution of the interpolationproblem (7.51). Then Q� WD � Q 2 Ms (by Theorem 7.13) and Q .zj/ D wj, sothat vj D Hzj.wj/ D Hzj.

Q .zj// D I Q�.zj/ by (7.16) for j D 1; : : : ; n C 1. Hence,by Proposition 7.32, . Q�/nC1 and �nC1 have the same moment sequence. But, sinceord .�/ D n, the measure �nC1 is determinate by Corollary 7.29(ii). Therefore,. Q�/nC1 D �nC1 which in turn implies that � Q � Q� D � � �˚ and hence Q D ˚

by Theorem 7.13. This shows that the interpolation problem (7.51) has a uniquesolution. It follows from Theorem 7.22 (iii)!(ii) that˚ is rational and deg.˚/ � n.

Now suppose that ˚ is rational and deg.˚/ � n. We proceed as in the precedingparagraph but in reverse order. Let � be a solution of the moment problem for �nC1and define Q� 2 MC.R/ by d Q� D QnC1

jD1 jx � zjj2d�. Then . Q�/nC1 D �. FromProposition 7.32, applied in the converse direction, it follows that IQ�.zj/ D I�.zj/ Dvj; j D 1; : : : ; n C 1; and Q� 2 Ms. Therefore, Q� D � Q for some Q 2 P byTheorem 7.13. Then we have vj D IQ�.zj/ D Hzj.

Q .zj// by (7.16), so that Q .zj/ DH�1

zj .vj/ D wj: Hence Q solves the interpolation problem (7.51) as well. Since ˚ isrational and deg.˚/ � n, the interpolation problem (7.51) has a unique solution by

7.8 Exercises 173

Theorem 7.22 (ii)!(iii). Thus, Q D ˚ which implies Q� D � and hence � D �nC1.This proves that �nC1 is determinate. Therefore, ord.�/ � n by Corollary 7.27.

In the paragraph before last it was shown that ord.�˚/ D n implies deg.˚/ � n:If we had deg.˚/ � n�1, then ord.�˚/ � n�1 by the preceding paragraph, whichis a contradiction. Thus we have proved that ord.��/ D deg.˚/. ut

7.8 Exercises

In this section, we assume that s is an indeterminate Hamburger momentsequence.

1. Let M be a compact subset of C. Then cM WD supz2M .P1

nD0 jpn.z/j2/1=2 < 1by Lemma 7.1. Show that jp.z/j � cMkpks for p 2 CŒx� and z 2 M.

2. Let z1; z2; z3; z4 2 C. Prove the following identities:

A.z1; z2/D.z3; z4/� B.z3; z2/C.z1; z4/C B.z3; z1/C.z2; z4/ D 0;A.z1; z2/C.z3; z4/C A.z3; z1/C.z2; z4/C A.z2; z3/C.z1; z4/ D 0;D.z1; z2/B.z3; z4/C D.z3; z1/B.z2; z4/C D.z2; z3/B.z1; z4/ D 0:

Hint: Verify the corresponding identities for Ak;Bk;Ck;Dk. Use Lemma 5.24.3. Let z;w 2 C. Show that

A.z;w/ D A.z/C.w/ � C.z/A.w/; B.z;w/ D A.z/D.w/� C.z/B.w/;

C.z;w/ D B.z/C.w/ �D.z/A.w/; D.z;w/ D B.z/D.w/� D.z/B.w/:

4. (Reproducing kernel) Recall that pk; k 2 N0; are the orthonormal polynomi-als.

a. Show that the series

K.z;w/ D1XkD0

pk.z/pk.w/; .z;w/ 2 C2;

converges uniformly on compact subsets of C2 to a holomorphic function K,called the reproducing kernel for s, such that D.z;w/ D .z � w/K.z;w/:

b. Show that for each representing measure � 2Ms and polynomial f 2 CŒx�,

ZR

K.z; x/f .x/d�.x/ D f .z/; z 2 C:

c. Show that the preceding equality remains valid for each holomorphicfunction f .z/ DP1

kD0 ckpk.z/, where .cn/ 2 l2.N0/.


5. Prove that the fractional linear transformation Hz;w defined by (7.10) maps

a. @Kw onto @Kz if Im z ¤ 0 and Im w ¤ 0;b. @Kw onto R if Im z D 0 and Im w ¤ 0;c. R onto @Kz if Im z ¤ 0 and Im w D 0;d. R onto R if Im z D Im w D 0:

6. Let Hz;w be the transformation defined by (7.10) and Hz WD Hz;0: Prove that

Hz;wHw;v D Hz;v ; H.z; z/ D I; .Hz;w/�1 D Hw;z; Hz;w D Hz.Hw/

�1; z;w; v 2 C:

Hint: Use (7.4) and (7.3).7. Show that for each n 2 N there is a continuum of measures � 2Ms of order

n.8. Suppose that � 2Ms has order n 2 N. Let p 2 RŒx� and define � 2MC.R/

by d� D .1C p.x/2/�1d�. What is the order of �?9. Show that each measure in Ms of finite order is an extreme point of the set Ms:

Hint: Use Proposition 1.21 and (for instance) Theorem 7.27 (iii).10. Consider the Nevanlinna–Pick interpolation problem in Theorem 7.22 and

sharpen the equivalence of (i) and (ii) therein: Show that 0 is an eigenvalueof multiplicity k if and only if the rational function ˚ has degree nC 1 � k.

11. Let ˚ D pq 2 P, where p; q 2 RŒx� have no common zeros. Suppose that the

support of the measure � in the representation (7.26) consists of n points.

a. Show that deg.˚/ D n and discuss the possible degrees of p and q.b. Express the polynomials p and q in terms of atoms and masses of � and of

the constant b in (7.26).

12. Collect characterizations of N-extremal solutions among all solutions (dense-ness of CŒx� in L2.R; �/, orthonormal basis fpk W k 2 N0g of L2.R; �/, valuesof the Stieltjes transform I�.z/ for z 2 CC, Nevanlinna parametrization, order).

13. Let � DP1kD1mkıxk be an N-extremal solution of s, with mk > 0 and xk 2 R

pairwise distinct. (Proofs of the following results can be found in [BC1].)

a. Show that the measure � WD ��PrkD1mkıxk is determinate for each r 2 N.

b. Let m0 > 0 and x0 2 R. Suppose that x0 ¤ xk for all k 2 N. Showthat the measure � D m0ıx0 C

P1kD2mkıxk is an N-extremal solution of

an indeterminate moment sequence.

7.9 Notes

Theorem 7.13 was proved in 1924 by R. Nevanlinna [Nv1]. Proposition 7.12 is dueto H. Hamburger [Hm] and R. Nevanlinna [Nv1]. The two-parameter Nevanlinnafunctions and fractional transformations appeared in [BCa2]. It is difficult todetermine explicit examples of Nevanlinna functions A;B;C;D. The first suchexamples were calculated in [ChiI], [IM], [BV], [CI].

7.9 Notes 175

The existence Theorem 7.20 on Nevanlinna–Pick interpolation is due to G. Pick[Pi] for finite sets Z and due to R. Nevanlinna [Nv2] for countable sets, see also [Ak,Theorem 3.3.3]. Operator-theoretic approaches to Nevanlinna–Pick interpolationare given in [SK] and [AM]. Characterizations of solutions of finite order such as inTheorem 7.27 were first obtained by H. Buchwalther and G. Cassier [BCa1]; furtherresults are in [Sim1] and [Ge].

There are important results about growth properties of Nevanlinna functions.Already M. Riesz [Rz2] had shown (see e.g. [Ak, p. 101]) that the entire functionsf D A;B;C;D are of minimal exponential type, that is, for each " > 0 there exists aK" > 0 such that jf .z/j � K"e"jzj for z 2 C. C. Berg and H.L. Petersen [BP] provedthat the four functions have the same order and type, called the order and type of theindeterminate moment sequence s. Further results are given in [BS2].

Chapter 8The Operator-Theoretic Approachto the Stieltjes Moment Problem

This chapter is devoted to a detailed study of Stieltjes moment problems by usingpositive self-adjoint extensions of positive symmetric operators on Hilbert spaces.

In Sect. 8.2 we rederive the existence theorem for the Stieltjes moment problemby operator-theoretic methods (Theorem 8.2). Since the Jacobi operator T for aStieltjes moment sequence s is positive, it has a largest positive self-adjoint exten-sion on Hs, the Friedrichs extension, and a smallest positive self-adjoint extension,the Krein extension. By the corresponding spectral measures this leads to twodistinguished solutions�F and�K of the Stieltjes moment problem for s. In Sect. 8.3we give an operator-theoretic characterization of Stieltjes determinacy by showingthat the Stieltjes moment problem is determinate if and only if the Jacobi operator Thas a unique positive self-adjoint extension on Hs (Theorem 8.7). The relationshipbetween Hamburger determinacy and Stieltjes determinacy is discussed. In Sect. 8.4we prove that for any other solution � of the Stieltjes moment problem the Stieltjestransforms satisfy I�F.x/ � I�.x/ � I�K .x/ for x < 0 (Theorem 8.18). Further,an approximation theorem for the Stieltjes transforms I�F.x/ and I�K .x/ is obtained(Theorem 8.16). Sections 8.5 and 8.6 develop the Nevanlinna parametrization ofsolutions (Theorem 8.24) and the Weyl circle description, respectively, for anindeterminate Stieltjes moment sequence.

8.1 Preliminaries on Quadratic Forms on Hilbert Spaces

In this short section we collect some facts on forms and positive self-adjointoperators that will be used in this chapter; all of them can be found in the book[Sm9].

Suppose that H is a Hilbert space. A positive quadratic form s on a linearsubspace DŒs� of H is a mapping sŒ�; �� W DŒs� � DŒs� ! C which is linear inthe first variable, antilinear in the second and satisfies sŒ'; '� � 0 for ' 2 DŒs�:


177

178 8 The Operator-Theoretic Approach to the Stieltjes Moment Problem

Such a form s is called closed if for each sequence .'n/n2N from DŒs� such thatlimn;k!1 sŒ'n�'k; 'n�'k� D 0 and limn!1 'n D ' in H for some ' 2 H we have' 2 DŒs� and limn!1 sŒ'n�'; 'n�'� D 0.

A symmetric operator T on H is said to be positive if hT'; 'i � 0 for ' 2 D.T/:For a positive symmetric operator T its greatest lower bound is the number

m.T/ WD sup f� 2 R W hT'; 'i � �h'; 'i for ' 2 D.T/g: (8.1)

Let A be a positive self-adjoint operator. Then the spectral measure EA issupported on Œm.A/;C1/ RC and A has a unique positive square root A1=2

given by A1=2 D R10 �1=2 dEA.�/. There exists a unique closed positive quadratic

form sA W

DŒsA� D D.A1=2/ and sAŒ'; � D hA1=2';A1=2 i for '; 2 DŒ sA�:

Conversely, for each densely defined closed positive quadratic form s there exists aunique positive self-adjoint operator A such that s D sA, see [Sm9, Theorem 10.17].

Let s1; s2 be positive quadratic forms on H. We define s1 � s2 if DŒs2� DŒs1�and s1Œ'; '� � s2Œ'; '� for all ' 2 DŒ s2�.

Let G1 and G2 be closed linear subspaces of H and let A1 and A2 be positiveself-adjoint operators on G1 and G2, respectively. We write A1 � A2 if sA1 � sA2 , orequivalently, D.A1=22 / D.A1=21 / and kA1=21 'k � kA1=22 'k for ' 2 D.A1=22 /.

Proposition 8.1 Let A1 and A2 be as above. Then A1 � A2 if and only if

.A2 � �I/�1 � .A1 � �I/�1

for one (then for all) � < 0. Here .Aj � �I/�1 denotes the operator of B.H/ whichis the inverse .Aj � �I/�1 on Gj and 0 on G?

j , j D 1; 2.Proof [Sm9, Corollary 10.13] in the case G1 D G2 D H. The general case is easilyobtained by minor modifications. ut

Now suppose that T is a densely defined positive symmetric operator on H. ThenT always has a positive self-ajoint extension on H. There is a largest positive self-adjoint extension, called the Friedrichs extension and denoted by TF , and a smallestpositive self-adjoint extension, called the Krein extension and denoted by TK , withrespect to the order relation “�” [Sm9, Corollary 13.15]. That is, if A is an arbitrarypositive self-adjoint extension of T on H, then we have

.TF C �I/�1 � .AC �I/�1 � .TK C �I/�1 for � > 0:

The Friedrichs extension is defined as follows. It can be shown that the positivequadratic form s defined by sŒ'; � D hT'; i, '; 2 DŒs� WD D.T/, has asmallest closed extension s. Since s is densely defined, closed, and positive, it isthe quadratic form of a unique positive self-adjoint operator. This operator is the

8.2 Existence of Solutions of the Stieltjes Moment Problem 179

Friedrichs extension TF, see [Sm9, Theorem 10.17]. From this construction of TF itfollows that T and TF have the same lower bounds, that is,

ms WD m.T/ D m.TF/: (8.2)

The Krein extension TK can be nicely described in the case when m.T/ > 0.Then we have (see [Sm9, formulas (14.67–68)])

D.TK/ D D.T/ PCN .T�/; TK.' C �/ D T' for ' 2 D.T/; � 2 N .T�/: (8.3)

If N .T�/ ¤ f0g, then 0 is an eigenvector of TK . Note that TF is invariant undertranslation, that is, .T � �I/F D TF � �I; but TK is not in general.

8.2 Existence of Solutions of the Stieltjes Moment Problem

In this short section we apply Hilbert space operator theory to solve the Stieltjesmoment problem. More precisely, we use the Friedrichs extension of a positivesymmetric operator and the spectral theorem for self-adjoint operators [Sm9]. Thefollowing theorem is the counterpart of Theorem 6.1 for the Stieltjes momentproblem.

Theorem 8.2 Suppose that s D .sn/n2N0 is a positive definite real sequence suchthat the sequence Es D .snC1/n2N0 is positive semidefinite. Then the Stieltjesmoment problem for s is solvable.

If A is a positive self-adjoint extension of the symmetric operator X on a possiblylarger Hilbert space G (that is, Hs G and X A) and EA is the spectral measureof A, then �A.�/DhEA.�/1; 1iG is a solution of the Stieltjes moment problem for s.Each solution of the Stieltjes moment problem for s is of this form.

Proof The proof follows the lines of the proof of Theorem 6.1 and we explain onlythe necessary modifications.

Let p.x/ D PnjD0 cjxj 2 CŒx�. Then xp.x/p.x/ D Pn

j;kD0 cjck xjCkC1. Therefore,since the sequence Es is positive semidefinite, we obtain

hXp; pis D L.xpp/ DnX

j;kD0cjck sjCkC1 � 0: (8.4)

This shows that the symmetric operator X is positive. The Friedrichs extension ofthe densely defined positive operator X is a positive self-adjoint extension. Hence Xhas at least one positive self-adjoint extension on Hs.

For any positive self-adjoint extension A of X, the spectral measure EA issupported on Œ0;C1/, so �A is a solution of the Stieltjes moment problem for s.


Conversely, if � is a solution of the Stieltjes moment problem for s, the self-adjoint operator A� from Proposition 6.2 is a positive self-adjoint extension of Xacting on the (possibly larger) Hilbert space L2.Œ0;C1/; �/. ut

Recall that the Jacobi operator T on l2.N0/ is unitarily equivalent to X. ThusTheorem 8.2 yields at once the following counterpart of Corollary 6.3.

Corollary 8.3 If s is as in Theorem 8.2, then the solutions of the Stieltjes momentproblem for s are precisely the measures of the form �B.�/ D s0hEB.�/e0; e0iF ,where B is a positive (!) self-adjoint extension of T on a possibly larger Hilbertspace F .

Suppose that s is a positive definite Stieltjes moment sequence. Then, by (8.4),the symmetric operator X Š T on Hs Š l2.N0/ is positive. Let ms denote thegreatest lower bound m.T/ of the operator T, see (8.1). From the extension theory ofpositive symmetric operators (see Sect. 8.1) it is known that T has a largest positiveself-adjoint extension on Hs, the Friedrichs extension TF , and a smallest positiveself-adjoint extension on Hs, the Krein extension TK . By (8.2), we have

ms D m.T/ D m.TF/ � 0: (8.5)

By Corollary 8.3, the spectral measures ETF and ETK give rise to solutions �F and�K , respectively, of the Stieltjes moment problem for s.

Definition 8.4 �F.�/ WD s0hETF .�/e0; e0i is the Friedrichs solution and �K.�/ WDs0hETK .�/e0; e0i is the Krein solution of the Stieltjes moment problem for s.

These two distinguished solutions �F and �K will play a crucial role in thischapter. Both solutions come from self-adjoint extensions of T on the Hilbert spaceHs Š l2.N0/, so they are von Neumann solutions according to Definition 6.4.

8.3 Determinacy of the Stieltjes Moment Problem

Suppose that s is a Stieltjes moment sequence and � is a solution of the Stieltjesmoment problem for s. If � is the only representing measure of s supportedon Œ0;C1/, then we say that s, and likewise �, is determinate or Stieltjesdeterminate if confusion can arise. But s may be indeterminate as a Hamburgermoment sequence, that is, s may have different representing measures on R (seeExample 8.11 below). Then the Stieltjes moment sequence s is called Ham-burger indeterminate. In order to distinguish these cases unambiguously we willspeak about Stieltjes determinacy and Hamburger determinacy of Stieltjes momentsequences in what follows. Obviously, if s is Hamburger determinate, it is alsoStieltjes determinate.

8.3 Determinacy of the Stieltjes Moment Problem 181

If s is Hamburger indeterminate, then �F and �K are N-extremal solutions of theHamburger moment problem for s according to Definition 7.8.

Let us begin with Hamburger indeterminate Stieltjes moment sequences.

Proposition 8.5 Suppose that s is a Stieltjes moment sequence which is Hamburgerindeterminate. Then the following are equivalent:

(i) X Š T has a unique positive self-adjoint extension on Hs Š l2.N0/.(ii) 0 is an eigenvalue of the Friedrichs extension TF of T.

(iii) ms D 0.Proof

(i)!(ii) The operator T1 from Theorem 6.23 is a positive self-adjoint extensionof T satisfying T1p0 D T�p0 D 0, that is, 0 is an eigenvalue of T1. Since T has aunique positive self-adjoint extension by (i), we have T1 D TF . This proves (ii).

(ii)!(iii) is trivial.(iii)!(i) Let A be an arbitrary positive self-adjoint extension of T on l2.N0/. The

Friedrichs extension TF is the largest positive self-adjoint extension of T, so thatA � TF . Hence, by Proposition 8.1,

.TF C I/�1 � .AC I/�1 � 1; (8.6)

where the second inequality holds because A is positive. By (iii) and (8.5) we havems D m.TF/ D 0. Therefore, k.TFCI/�1k D 1 and hence k.ACI/�1k D 1 by (8.6).Since A � 0 and A has a discrete spectrum by Theorem 7.7(i), it follows from theequality k.AC I/�1k D 1 that 0 is an eigenvalue of A. But 0 is also in the spectrumof TF , because m.TF/ D 0. Therefore, from Theorem 7.7(ii) it follows that A D TF .This shows that TF is the unique self-adjoint extension of T on l2.N0/. utCorollary 8.6 Let s be a Stieltjes moment sequence which is Hamburger indeter-minate. Then the operator T1 from Theorem 6.23 is the Krein extension TK of Tand

D.TK/ D D.T1/ D D.T/ PCC � p0 D D.T/ PCN .T�/; (8.7)

TK.' C �p0/ D T' for ' 2 D.T/; � 2 C: (8.8)

Proof Because s is Hamburger indeterminate, we have p0 2 l2.N0/ by Theo-rem 6.16 and hence N .T�/ D C � p0 by Proposition 6.6(i).

First let ms D 0. Then, by Proposition 8.5, T has only one positive self-adjointextension on l2.N0/. Since TK (by definition) and T1 (by Theorem 6.23) are suchextensions, T1 D TK and (6.16) implies (8.7) and (8.8).

Now suppose that ms ¤ 0. Then ms > 0 and hence D.TK/ D D.T/ PCN .T�/ by(8.3). Therefore, since N .T�/ D C � p0, we obtain D.T1/ D D.TK/. But T1 andTK are restrictions of T�, so that T1 D TK and (6.16) yields (8.7) and (8.8). ut


The main result in this section is the following operator-theoretic characterizationof Stieltjes determinacy. It is the counterpart of the corresponding result (Theo-rem 6.10) for the Hamburger moment problem.

Theorem 8.7 Suppose that s is a positive definite Stieltjes moment sequence. Thens is Stieltjes determinate if and only if the symmetric operator X, or equivalently theJacobi operator T, has a unique positive self-adjoint extension on the Hilbert spaceHs Š l2.N0/.

Proof First we assume that X has two different positive self-adjoint extensions, sayA and B, on Hs. We repeat the reasoning from the proof of Theorem 6.10. Then�A.�/DhEA.�/1; 1i and �B.�/DhEB.�/1; 1i are representing measures for s. They aresupported on Œ0;C1/, because A and B are positive. If �A were equal to �B, thenwe would have h.A�zI/�11; 1i D h.B�zI/�11; 1i for z 2 CnR by the functionalcalculus of self-adjoint operators. This contradicts Lemma 6.8. Hence �A ¤ �B, sos is Stieltjes indeterminate.

Now we assume that X has a unique positive self-adjoint extension on Hs. If s isHamburger determinate, it is Stieltjes determinate and we are finished. Suppose nowthat s is Hamburger indeterminate. Then �F is N-extremal and 0 is an eigenvalueof TF by Proposition 8.5. Since the multiplication operator X on Hs and the Jacobioperator T on l2.N0/ are unitarily equivalent, so are their Friedrichs extensions XF

and TF , and we have �F.�/ D hEXF.�/1; 1is. Then 0 is an eigenvalue of XF. Letf 2 Hs be a corresponding unit eigenvector. From the definition of the Friedrichsextension it follows that there exists a sequence . fn/n2N from D.X/ D CŒx� suchthat limn fn D f in Hs Š L2.RC; �F/ and limn hXfn; fnis D hXFf ; f is D 0.

Let � be an arbitrary solution of the Stieltjes moment problem for s. Since wehave fn ! f in Hs, . fn/n2N is a Cauchy sequence in .CŒx�; k�ks/ and so in L2.RC; �/by Proposition 6.2. Hence fn ! g in L2.RC; �/ for some g 2 L2.RC; �/. Clearly,kgkL2.RC;�/

D 1, since k fnkL2.RC;�/D k fnks ! k fks D 1. Then

Z 1

0

jpxfnj2 d� DZ 1

0

xfn fn d� D Ls.xfn fn / D hXfn; fnis ! hXFf ; f is D 0:

Therefore, for each function ' 2 Cc.RCIR/ we obtain

Z 1

0

pxfn ' d� D

Z 1

0

fnpx ' d�! 0 D

Z 1

0

gpx ' d�:

This implies that g.x/ D 0 �-a.e. on .0;C1/. Since fn ! f in L2.RC; �F/, wehave in particular f .x/ D 0 �F-a.e. on .0;C1/. (This also follows from the factthat XFf D 0.) Thus, since g 2 L2.RC; �/ and f 2 L2.RC; �F/ are unit vectors, weget

�.f0g/jg.0/j2 D �F.f0g/j f .0/j2 D 1: (8.9)

8.3 Determinacy of the Stieltjes Moment Problem 183

Further, we have

Z 1

0

fn d� D Ls. fn/ DZ 1

0

fn d�F !Z 1

0

g d� DZ 1

0

f d�F:

Since g.x/ D 0 �-a.e. and f .x/ D 0 �F-a.e. on .0;C1/, the latter equality yields

�.f0g/g.0/D �F.f0g/f .0/: (8.10)

Combining (8.9) and (8.10) we obtain �.f0g/ D �F.f0g/ > 0. Because �F is a vonNeumann solution of the indeterminate Hamburger moment sequence s, it followsfrom Corollary 7.17 that � D �F . This proves that s is Stieltjes determinate. ut

We close this section by deriving three useful corollaries. An immediate conse-quence of Theorem 8.7 and Proposition 8.5 is the following.

Corollary 8.8 Let s be a Stieltjes moment sequence which is Hamburger indeter-minate. The following are equivalent:

(i) s is Stieltjes determinate.(ii) 0 is an eigenvalue of the Friedrichs extension TF of T.

(iii) ms D 0.Corollary 8.9 Let s be a determinate Stieltjes moment sequence with representingmeasure �. If �.f0g/ D 0, then s is Hamburger determinate.Proof Since s is Stieltjes determinate, � D �F . The multiplication operator A� bythe variable x on L2.Œ0;C1/; �/ and the Friedrichs extension TF are positive self-adjoint extensions of X Š T on Hs Š L2.Œ0;C1/; �/ D L2.Œ0;C1/; �F/. HenceA� D TF by Theorem 8.7. Since �.f0g/ D 0, 0 is not an eigenvalue of A� D TF .Hence s cannot be Hamburger indeterminate by Corollary 8.8. utCorollary 8.10 Suppose that s is an indeterminate Stieltjes moment sequence. Thenm.TF/ > 0, supp�F Œm.TF/;C1/, and 0 is in the resolvent set of the Friedrichsextension TF, that is, .TF/�1 2 B.Hs/.

Proof Since s is Stieltjes indeterminate, it is Hamburger indeterminate and T hasat least two different positive self-adjoint extensions on l2.N0/ by Theorem 8.7.Therefore, ms D m.TF/ > 0 by Proposition 8.5 and (8.5). From the theory of self-adjoint operators it follows that the spectrum of TF , hence the support of the measure�F , is contained in Œm.TF/;C1/, so that 0 is in the resolvent set of TF. utExample 8.11 (A determinate Stieltjes moment sequence that is Hamburger inde-terminate) Let s be an indeterminate Stieltjes moment sequence. Then ms Dm.TF/ > 0 by Corollary 8.10. Let Qs D .Qsn/n2N0 denote the shifted sequence of sby �ms, that is, Qsn D Pn

kD0�nk

�.�ms/

ksn�k for n 2 N0; see Exercise 6.5. Then Qs isHamburger indeterminate (because s is Hamburger indeterminate) and mQs D 0. ByCorollary 8.8, mQs D 0 implies that Qs is Stieltjes determinate. ı


8.4 Friedrichs and Krein Approximants

In this section we suppose that s is a positive definite Stieltjes moment sequence.We develop two sequences of matrices and two related sequences of quotients ofpolynomials as approximants for the Friedrichs and Krein extensions and as usefultools in the proofs of our main results (Theorems 8.16 and 8.18).

The truncated Jacobi matrix (6.21) is called the Friedrichs approximant anddenoted by AŒn�F , that is, we set

AŒn�F WD Jn D

0BBBBBBB@

b0 a0 0 : : : 0 0 0

a0 b1 a1 : : : 0 0 0

0 a1 b2 : : : 0 0 0

: : : : : : : : : : : : : : : : : : : : :

0 0 0 : : : an�3 bn�2 an�20 0 0 : : : 0 an�2 bn�1

1CCCCCCCA; n 2 N: (8.11)

The Krein approximant AŒn�K is defined by

AŒn�K D

0BBBBBBB@

b0 a0 0 : : : 0 0

a0 b1 a1 : : : 0 0

0 a1 b2 : : : 0 0

: : : : : : : : : : : : : : :

0 0 0 : : : bn�2 an�20 0 0 : : : an�2 bn�1 � ˛n�1

1CCCCCCCA; n 2 N; (8.12)

where ˛n�1 is chosen according to the following lemma.

Lemma 8.12 There is a unique positive number ˛n�1 such that AŒn�K has theeigenvalue zero. A corresponding eigenvector is . p0.0/; : : : ; pn�1.0//. The matrixAŒn�K is positive semidefinite, pn�1.0/ ¤ 0, and we have

.bn�1 � ˛n�1/pn�1.0/C an�2pn�2.0/ D 0; (8.13)

˛n�1 D� an�1pn.0/

pn�1.0/; (8.14)

.bn � ˛n/˛n�1 � a2n�1 D 0: (8.15)

Proof The assertion is obvious for n D 1, so we assume that n � 2. Fix ˛n�1 andy D . y0; : : : ; yn�1/ 2 Cn with y0 WD p0.0/. Let us consider the equation AŒn�K y D 0.For the first n�1 components of y this is equivalent to the recurrence relations (5.9)for x D 0 with the same intial data. Therefore, yk D pk.0/ for k D 0; : : : ; n � 1. Forthe n-th component this is precisely equation (8.13).

8.4 Friedrichs and Krein Approximants 185

The sequence s is positive definite, hence is AŒn�F . Therefore, pn�1.0/ ¤ 0, because

otherwise AŒn�F would have the eigenvalue zero. Since pn�1.0/ ¤ 0, (8.13) has aunique solution ˛n�1. Let Cn denote the matrix with entry one at the right lowercorner and zero otherwise. Then AŒn�F D AŒn�K C ˛n�1Cn by definition. If ˛n�1 � 0,

then 0 � AŒn�F � AŒn�K and hence AŒn�F would not be positive definite. Hence ˛n�1 > 0.The recurrence relation (5.9) gives an�1pn.0/Cbn�1pn�1.0/Can�2pn�2.0/ D 0.

Combined with (8.13) this yields (8.14). Replacing now n�1 by n in Eq. (8.13) andcomparing this with (8.14) we obtain (8.15). utLemma 8.13 For n 2 N0 we define

Mn.z/ D Pn.z/ � Pn.0/

Pn�1.0/Pn�1.z/; Nn.z/ D Qn.z/� Pn.0/

Pn�1.0/Qn�1.z/:

Then

h.AŒn�K � zI/�1e0; e0i D � Nn.z/

Mn.z/; z 2 .AŒn�K /: (8.16)

Proof In this proof we use Lemmas 6.27 and 6.28. Let BŒn�K be the matrix obtained

from AŒn�K by removing the first row and the first column. By developing thedeterminant and using Lemma 6.27(i) we get

det .zI � AŒn�K / D det .zI � AŒn�F /� ˛n�1 det .zI � AŒn�1�F / D Pn.z/� ˛n�1Pn�1.z/:

Since the matrix AŒn�K has the eigenvalue 0 by Lemma 8.12, we have det AŒn�K D 0.Hence Pn.0/ D ˛n�1Pn�1.0/ and

det .zI � AŒn�K / D Pn.z/� Pn.0/Pn�1.0/�1Pn�1.z/ D Mn.z/: (8.17)

Similarly, applying Lemma 6.28(i) we derive

det .zI � BŒn�K / D det .zI � BŒn�F / � ˛n�1 det .zI � BŒn�1�F /

D Qn.z/ � ˛n�1Qn�1.z/ D QnC1.z/� Pn.0/Pn�1.0/�1Qn�1.z/ D Nn.z/:

As in the proof of Lemma 6.28 we use Cramer’s rule and obtain

h.AŒn�K � zI/�1e0; e0i D det .BŒn�K � zI/

det .AŒn�K � zI/D �det .zI � BŒn�K /

det .zI � AŒn�K /D � Nn.z/

Mn.z/

for z in the resolvent set .AŒn�K /. This completes the proof of Lemma 8.13. ut


Lemma 8.14 For x < ms and n 2 N0, we have

h.AŒn�F � xI/�1e0; e0i � h.AŒnC1�F � xI/�1e0; e0i; (8.18)

limn!C1h.A

Œn�F � xI/�1e0; e0i D h.TF � xI/�1e0; e0i : (8.19)

Proof In this proof we use some facts on forms and self-adjoint operators, seeSect. 8.1. Let sn denote the positive quadratic form defined by

sn. f ; g/ D hAŒn�F f ; gi; f ; g 2 DŒ sn� WD f. f0; : : : ; fn�1; 0; 0; : : : / W fj 2 Cg:

Then DŒsn� DŒsnC1� and sn. f ; f / D snC1. f ; f / for f 2 DŒsn�. By the definition ofthe order relation of forms this means that snC1 � sn. Therefore, by Proposition 8.1,

.AŒn�F � xI/�1 � .AŒnC1�F � xI/�1: (8.20)

(By the convention in Proposition 8.1 the resolvents are defined to be 0 on theorthogonal complements of DŒsn� and DŒsnC1� in l2.N0/.) Clearly, (8.20) implies(8.18).

Since ms > x, the sequence ..AŒn�F � xI/�1/n2N of bounded positive self-adjointoperators on l2.N0/ is monotonically increasing by (8.20) and bounded from aboveby .ms�x/�1I (since AŒn�F � msI). Hence it converges strongly to a bounded positiveself-adjoint operator S such that S � .ms�x/�1I.

Let f 2 N .S/. Then 0 D hSf ; f i � h.AŒn�F � xI/�1f ; f i � 0: Since AŒn�F � msI;

this implies .AŒn�F � xI/�1f D 0 for all n 2 N. Hence f 2 \n DŒsn�? D f0g, so thatf D 0. Thus, S has a trivial kernel. Therefore, since S � .ms�x/�1I, it follows thatA WD S�1 C xI is a positive self-adjoint operator on l2.N0/ and

.AŒn�F � xI/�1 � S D .A � xI/�1; n 2 N: (8.21)

We prove that A D TF . Let sA denote the positive quadratic form associated withA: By definition the Friedrichs extension TF of T is the positive self-adjoint operatorassociated with the closure s1 of the quadratic form defined by

s1. f ; g/ D hTf ; gi; f ; g 2 DŒs1� D d:

By (8.21) and Proposition 8.1, sA � sn. Therefore, DŒsn� DŒsA� for all n 2 N andhence DŒs1� D [nDŒsn� DŒsA�. Further, sn. f ; f / D s1. f ; f / for f 2 DŒsn� and

sA. f ; f / � limn

sn. f ; f / D s1. f ; f / for f 2 DŒs1�:


The preceding facts show that sA � s1 � sn. Since sn is closable, sA � s1 � sn.Applying again Proposition 8.1 we conclude that

.AŒn�F � xI/�1 � .TF � xI/�1 � .A � xI/�1 D S:

But S is the strong limit of the sequence ..AŒn�F � xI/�1/n2N. Hence the latter impliesthat .TF � xI/�1 D .A � xI/�1 D S: Therefore, TF D A. Thus, for f 2 l2.N0/,

.TF � xI/�1f D .A � xI/�1f D Sf D limn.AŒn�F � xI/�1f ;

which in turn yields (8.19). utNow we treat the Krein approximants. We extend AŒn�K to a positive self-adjoint

finite rank operator T Œn�K on l2.N0/ by filling up the matrix with zeros, that is,

T Œn�K WD AŒn�K 0

0 0

!:

Then it is obvious that

h.T Œn�K � zI/�1e0; e0i D h.AŒn�K � zI/�1e0; e0i; z 2 .T Œn�K / D .AŒn�K /: (8.22)

The following Eq. (8.24) says that the self-adjoint operator TK is the strong resolventlimit of the sequence .T Œn�K /n2N.

Lemma 8.15 For x < 0, n 2 N0, and f 2 l2.N0/,

h.T ŒnC1�K � xI/�1e0; e0i � h.T Œn�K � xI/�1e0; e0i; (8.23)

limn!C1.T

Œn�K � xI/�1f D .TK � xI/�1f : (8.24)

Proof The nonzero part of the matrix T ŒnC1�K � T Œn�K is the block matrix

Dn D�˛n�1 an�1an�1 bn � ˛n

�:

By (8.15), detDn D ˛n�1.bn � ˛n/ � a2n�1 D 0. In particular, ˛n�1.bn � ˛n/ � 0.Hence bn�˛n � 0, since ˛n�1 > 0 by Lemma 8.12, and TrDn D bn�˛nCan�1 � 0,since an�1 > 0. Since det Dn D 0 and TrDn � 0, it follows that Dn � 0 and henceT ŒnC1�K � T Œn�K � 0. Therefore .T ŒnC1�

K � xI/�1 � .T Œn�K � xI/�1; which implies (8.23).

Now we prove (8.24). Since k.T Œn�K � xI/�1k � jxj�1 for all n 2 N, it is easilyshown that the set of f 2 l2.N0/ for which (8.24) holds is closed. Further, (8.24) isvalid for f 2 .T � xI/D.T/. Indeed, then T Œn�K f D Tf D TKf for some n 2 N and

hence .T Œn�K � xI/�1f D .TK � xI/�1f .


First assume that s is Hamburger determinate. Then T is essentially self-adjointby Theorem 6.10. Hence T D TK . Since T � 0 and x < 0, .T � xI/D.T/ is dense inl2.N0/ by Proposition A.42(iv). Thus (8.24) holds on l2.N0/ as noted above.

Now suppose s is Hamburger indeterminate. Then we have D.TK/DD.T/CC �p0and TKp0 D 0 by (8.7). From .TK � xI/p0 D �xp0 we get .TK � xI/�1p0 D �x�1p0.Put p

Œn�0 WD . p0.0/; : : : ; pn.0/; 0; : : : /. Since T Œn�K p

Œn�0 D 0 by Lemma 8.12, we have

.T Œn�K � xI/�1pŒn�0 D �x�1pŒn�0 . Then, as n!1,

k.T Œn�K � xI/�1p0 � .TK � xI/�1p0k D k.T Œn�K � xI/�1p0 � x�1p0kD k.T Œn�K � xI/�1.p0 � p

Œn�0 /C x�1.pŒn�0 � p0/k � 2jxj�1k.pŒn�0 � p0/k ! 0:

This proves (8.24) for f D p0. Because TK � 0 is self-adjoint and x < 0, it followsfrom Proposition A.42(iv) and (8.7) that

.TK � xI/D.TK/ D .T � xI/D.T/CC � p0 D .T � xI/D.T/C C � p0 D l2.N0/:

Therefore, since (8.24) is valid for f 2 .T � xI/D.T/ and f D p0 as shown above,(8.24) holds for all f 2 l2.N0/. ut

The zeros of Pn.x/ are contained in Œms;C1/ by Proposition 5.28 (or byLemma 6.27(i)). Hence Pn.x/ ¤ 0 for x < ms. Moreover, Mn.x/ ¤ 0 for x < 0 by(8.17).

Putting the preceding together we obtain our main approximation result.

Theorem 8.16 Suppose that s is a positive definite Stieltjes moment sequence.For any x < ms the sequence

� � Qn.x/Pn.x/

�n2N0

is bounded increasing and

�s0 limn!1

Qn.x/

Pn.x/DZ 1

0

d�F. y/

y � x: (8.25)

For x < 0 the sequence�� Nn.x/

Mn.x/

�n2N0

is bounded decreasing and

�s0 limn!1

Nn.x/

Mn.x/DZ 1

0

d�K. y/

y � x: (8.26)

Proof Recall that AŒn�F acting as an operator on Cn is just the operator Jn in(6.24). Combining (6.24), (8.18), and (8.19) it follows that .�Qn.x/

Pn.x//n2N0 is a

bounded increasing sequence converging to h.TF � xI/�1e0; e0i. Since �F.�/ Ds0hETF .�/e0; e0i by Definition 8.4, we have s0h.TF � xI/�1e0; e0i D

R10

d�F. y/y�x by

the functional calculus, whence (8.25) follows.Similarly, we conclude from (8.16), (8.22), (8.23), and (8.24) that the sequence

.� Nn.x/Mn.x/

/ is bounded, decreasing, and that it converges to h.TK � xI/�1e0; e0i.Combined with s0h.TK � xI/�1e0; e0i D

R10

d�K . y/y�x this yields (8.26). ut


Remark 8.17 The assertion of Theorem 8.16 also holds for a determinate positivedefinite Stieltjes moment sequence s. In this case � WD �K D �F is the uniquesolution of the Stieltjes moment problem for s and it follows from Theorem 8.16that

��s0 Qn.x/Pn.x/

�n2N0

and��s0 Nn.x/

Mn.x/

�n2N0

are monotone sequences converging from

below resp. from above to the Stieltjes transform I�.x/ DR10. y � x/�1 d�. y/ of

the representing measure � for x < 0. ıOur second main result in this section is the following theorem.

Theorem 8.18 Suppose that s is an indeterminate Stieltjes moment sequence. If �is an arbitrary solution of the Stieltjes moment problem for s, then

Z 1

0

d�F. y/

y � x�Z 1

0

d�. y/

y � x�Z 1

0

d�K. y/

y � xfor x < 0; (8.27)

Z 1

0

d�. y/

y � x<

Z 1

0

d�K. y/

y � xif � ¤ �K ; x < 0; (8.28)

Z 1

0

d�F. y/

y � x<

Z 1

0

d�. y/

y � xif � ¤ �F; x � 0: (8.29)

Remark 8.19 It should be emphasized that both inequalities (8.28) and (8.29) arestrict and that (8.29) also holds for x D 0. In this case the integral

R10

y�1d�. y/in (8.29) can be infinite, while

R10

y�1d�F. y/ is always finite, since supp�F Œm.TF/;C1/ and m.TF/ > 0 by Corollary 8.10. ıProof The proofs of the two strict inequalities (8.28) and (8.29) will be given at theend of the proof of Theorem 8.24 below. Here we only prove the inequalities (8.27).

Since an indeterminate Stieltjes moment sequence is obviously positive definite,Theorem 8.16 applies and by (8.25) and (8.26) it suffices to show that

�s0 Qn.x/

Pn.x/�Z 1

0

d�. y/

y � x� �s0 Nn.x/

Mn.x/for x < 0: (8.30)

Recall that Hs is a subspace of L2.R; �/ by Proposition 6.2. Now we derive

0 �Z 1

0

Pn. y/2

y � xd�. y/

DZ 1

0

Pn. y/Pn. y/� Pn.x/

y � xd�. y/C Pn.x/

Z 1

0

Pn. y/

y � xd�. y/

.1/D Pn.x/Z 1

0

Pn. y/

y � xd�. y/

D Pn.x/Z 1

0

Pn. y/ � Pn.x/

y � xd�. y/C Pn.x/

2

Z 1

0

d�. y/

y � x


.2/D Pn.x/ Ls;y

�Pn. y/ � Pn.x/

y � x

�C Pn.x/

2I�.x/

.3/D Pn.x/ s0Qn.x/C Pn.x/2I�.x/:

Dividing this inequality by Pn.x/2 leads to the left inequality of (8.30).Let us explain why the three equalities (1)–(3) of the preceding derivation hold.

Since .Pn. y/� Pn.x//=. y� x/ is a polynomial in y of degree less than degPn D n,it is orthogonal to Pn. y/ in Hs and so in L2.RC; �/. Thus,

Z 1

0

Pn. y/Pn. y/� Pn.x/

y � xd�. y/ D 0:

This has been used in (1). Because � is a solution of the Stieltjes moment problemfor s, equality (2) holds. Finally, (3) follows from (5.34).

Now we turn to the right inequality of (8.30). Since Mn.0/ D 0 by the definitionof Mn, Mn. y/y�1 is a polynomial of degree n � 1 in y. Hence

Mn. y/y�1 �Mn.x/x�1

y � x

is a polynomial of degree at most n � 2 in y. Therefore, it is orthogonal to Pn�1. y/and Pn. y/ and hence to Mn. y/ in Hs and so in L2.R; �/. This gives the equality(4) below. Further, from the definitions of Nn and Mn (see Lemma 8.13) and the

equation s0Qn.x/ D Ls;y�Pn. y/�Pn.x/

y�x

by (5.34) it follows that

Z 1

0

Mn. y/�Mn.x/

y � xd�. y/ D Ls;y

�Mn. y/�Mn.x/

y � x

�D s0 Nn.x/:

This relation is inserted in equality (5) below. Using the preceding facts we derive

0 �Z 1

0

Mn. y/2

. y � x/yd�. y/

DZ 1

0

Mn. y/Mn. y/y�1 �Mn.x/x�1

y � xd�. y/CMn.x/x

�1Z 1

0

Mn. y/

y � xd�. y/

.4/D Mn.x/x�1Z 1

0

Mn. y/

y � xd�. y/

D Mn.x/x�1Z 1

0

Mn. y/�Mn.x/

y � xd�. y/CMn.x/

2x�1Z 1

0

d�. y/

y � x

.5/D Mn.x/x�1 s0 Nn.x/CMn.x/

2x�1I�. y/:

Dividing now by Mn.x/2x�1 < 0 yields the right inequality of (8.30). ut


By Theorem 8.16, the sequence� � Qn.x/

Pn.x/

�n2N0

is monotonically increasing and

the sequence�x Nn.x/Mn.x/

�n2N0

is monotonically decreasing for x < 0. Letting x! �0 it

follows that�� Qn.0/

Pn.0/

�is also monotonically increasing and

� Nn.0/

M0

n.0/

�is monotonically

decreasing. (Note that Pn.0/ ¤ 0 by Lemma 8.12 and limx!�0 xNn.x/Mn.x/

D Nn.0/

M0

n.0/; since

Mn.0/ D 0.) Hence the limits �s and ˇs of these sequences exist. They enter into thefollowing corollary.

Corollary 8.20 For each positive definite Stieltjes moment sequence s we have

�s WD �s0 limn!1

Qn.0/

Pn.0/DZ 1

0

d�F. y/

y; ˇs WD s0 lim

n!1Nn.0/

M0n.0/D �K.f0g/:

(8.31)

Proof From Theorem 8.16 we obtain

�s0 Qn.x/

Pn.x/�Z 1

0

d�F. y/

y � x�Z 1

0

d�F. y/

y

for x < 0. Letting x! �0 and then n!1 we get �s �Ry�1d�F. y/.

We prove the converse inequality. Recall that Jn D AŒn�F by (8.11). Hence, by

Lemma 6.28 we have h.AŒn�F � xI/�1e0; e0i D �Qn.x/Pn.x/

for x < 0. Since AŒn�F � 0, theleft-hand side increases as x! �0 and we obtain

�s0 Qn.x/

Pn.x/� �s0 Qn.0/

Pn.0/� �s; x < 0:

Letting n!1 and using (8.25) this yieldsR. y� x/�1d�F. y/ � �s. Passing to the

limit x ! �0 by using Lebesgue’s monotone convergence theorem we concludethat

Ry�1d�F. y/ � �s. This completes the proof of the first equality in (8.31).

By Theorem 8.16, the sequence .�s0 Nn.x/Mn.x/

/ converges toR. y�x/�1 d�K. y/ from

above for x < 0. Therefore, multiplying by �x > 0; we obtain

�K.f0g/ �Z 1

0

.�x/d�K. y/

y � x� s0

xNn.x/

Mn.x/:

Passing to the limits x! �0 and then n!1 yields �K.f0g/ � ˇs.Conversely, the expression h.�x/.AŒn�K � xI/�1e0; e0i D xNn.x/

Mn.x/(by Lemma 8.13)

decreases as x! �0, since AŒn�K � 0. Therefore, for x < 0,

s0xNn.x/

Mn.x/� s0

Nn.0/

M0n.0/

:


Now we take the limit n!1. Because of (8.26) we then obtain

Z 1

0

.�x/d�K. y/

y � x� ˇs; x < 0: (8.32)

Since j �xy�x j � 1 for y � 0 and limx!�0. �x

y�x / is the characteristic function of thepoint 0, Lebesgue’s dominated convergence theorem applies for the limit x ! �0in (8.32) and yields �K.f0g/ � ˇs. This proves the second equality in (8.31). ut

Note that �s can be infinite if the Stieltjes moment sequence s is determinate.However, if s is indeterminate, then �s is finite (by Corollary 8.21). In this case thenumber �s enters into the Nevanlinna parametrization given in Theorem 8.24 below.

Corollary 8.21 Let s be a positive definite Stieltjes moment sequence. Then s isStieltjes indeterminate if and only if �s <1 and ˇs ¤ 0:Proof Let s be Stieltjes indeterminate. Then m.TF/ > 0 by Corollary 8.10 andsupp�F Œm.TF/;C1/, so that �s D

R10

y�1d�F. y/ < 1. Obviously, s is alsoHamburger indeterminate. Hence p0 2 D.TK/ by Corollary 8.6. Since TKp0 DT�p0 D 0, 0 is an eigenvalue of TK and hence ˇs D �K.f0g/ D kp0k�2 ¤ 0 byformula (7.7).

Conversely, assume that �s < 1 and ˇs ¤ 0. Since thenR10

y�1d�F. y/ < 1and �K.f0g/ ¤ 0, we have �F ¤ �K . Hence s is Stieltjes indeterminate. ut

The next corollary uses the inquality (8.29) which will be proved only in the nextsection. It can be used to construct determinate moment sequences of “fast growth”.

Corollary 8.22 Suppose that s D .sn/n2N0 is an indeterminate Stieltjes momentsequence. Set c WD R1

0y�1d�F. y/, Qs0 D 1, and Qsn D c�1sn�1 for n 2 N. Then

Qs D .Qsn/n2N0 is a Stieltjes moment sequence which is Hamburger determinate.

Proof First note that c 2 .0;C1/, since supp�F Œm.TF/;C1/ and m.TF/ > 0

by Corollary 8.10. Clearly, the measure �0 given by d�0 D c�1y�1d�F is supportedon RC and has the moments Qsn. Hence Qs is a Stieltjes moment sequence.

Let � be a solution of the Stieltjes moment problem for Qs. Then the measure �given by d�. y/ WD cyd�. y/ has the moment sequence s, is supported on RC, and

Z 1

0

y�1d�. y/ D cZ 1

0

d�. y/ D cQs0 D c DZ 1

0

y�1d�F. y/:

Therefore, it follows from statement (8.29) in Theorem 8.18, applied with x D 0,that � D �F . This implies � D �0. Hence Qs is a determinate Stieltjes momentsequence with representing measure �0. Since �0.f0g/ D 0, Qs is also Hamburgerdeterminate by Corollary 8.9. ut

8.5 Nevanlinna Parametrization for the Indeterminate Stieltjes Moment Problem 193

8.5 Nevanlinna Parametrization for the IndeterminateStieltjes Moment Problem

For � 2 R let P� denote the set of Pick functions ˚ 2 P which are holomorphicon CnRC and map .�1; 0/ into Œ�;1/. Note that all constant functions equal to anumber t 2 Œ�;1/ are contained in P� . Set P� WD P� [ f1g.Proposition 8.23 A function ˚ belongs to the class P� if and only if

˚.z/ D ˛ CZ 1

0

d�.x/

x � z; z 2 CnŒ�;1/; (8.33)

where ˛ � � and � is a positive Borel measure on RC satisfyingR10

d�.x/xC1 <1.

Proof Clearly, any function ˚ of the form (8.33) is holomorphic on CnRC. Sincea � � and

R10.x � z/d�.x/ � 0 for z < 0, ˚ maps .�1; 0/ into Œ�;1/, so that

˚ 2 P� .Conversely, let˚ 2 P� . Then˚ 2 P, so by formula (A.5) it has a representation

˚.z/ D aC bzCZR

�1

x � z� x

1C x2

�d�.x/; z 2 CnR; (8.34)

where a; b 2 R, b � 0, and � is a positive measure such thatR.1Cx2/�1d�.x/ <1.

Since ˚ is holomorphic on CnRC, Proposition A.15 implies that supp � RC.We consider the limit z!�1 in (8.34). The integrand converges monotonically

decreasing on RC to the function � x1Cx2

. If b > 0, it follows from Lebesgue’smonotone convergence theorem that ˚.z/ ! �1 as z ! �1. This contradicts˚ 2 P� . Thus b D 0. Applying the limit z ! �1 in (8.34) once more, ˚ 2 P�

implies that c WD R10

x.1 C x2/�1d�.x/ < 1 and ˚.z/ ! a � c. This yieldsR10.1 C x/�1d�.x/ < 1 and ˛ WD a � c � �; since ˚ 2 P� . Then (8.34) gives

(8.33). utSuppose that s is an indeterminate Stieltjes moment sequence. Let �s denote the

positive real number (see Corollary 8.21) defined by (8.31). Recall that m.TF/ > 0,supp�F Œm.TF/;C1/, and .TF/�1 2 B.Hs/ by Corollary 8.10. If ETF denotesthe spectral measure of the self-adjoint operator TF , the functional calculus yields

�s �Z 1

0

y�1 d�F. y/ D s0

Z 1

0

y�1 dhETF. y/e0; e0i D s0 h.TF/�1e0; e0i:(8.35)

The positive number �s is called the Friedrichs parameter of s.The following theorem is the counterpart of Nevanlinna’s Theorem 7.13 for the

Stieltjes moment problem.


Theorem 8.24 Suppose that s is an indeterminate Stieltjes moment sequence. Thenthere is a one-to-one correspondence between functions � 2 P�s

and solutions � ofthe Stieltjes moment problem for s given by

I�� .z/ �Z 1

0

1

x � zd��.x/ D �A.z/C �.z/C.z/

B.z/C �.z/D.z/ � Hz.�.z//; z 2 CC:

(8.36)

Proof Since s is Stieltjes indeterminate, it is Hamburger indeterminate. ThusTheorem 7.13 applies and gives a one-to-one correspondence �˚ $ ˚ betweensolutions �˚ of the Hamburger moment problem for s and functions ˚ 2 P.Therefore, it suffices to show that supp�˚ RC if and only if ˚ 2 P�s

.

Recall that the solutions �t for t 2 R P correspond to the self-adjointoperators Tt from Theorem 6.23. Since TK D T1 by Corollary 8.6, the assertionholds for � D �K and ˚ D 1. Further, s0 hT�1

t e0; e0i D t for t 2 R byLemma 6.24. Hence Eq. (8.35) implies that TF D T�s , so that the assertion is alsovalid for ˚ D �s.

Let us fix x 2 R and consider the fractional linear transformation (see (7.10))

Hx.t/ WD �A.x/C tC.x/

B.x/C tD.x/:

Since A.x/;B.x/;C.x/;D.x/ 2 R and A.x/D.x/ � B.x/C.x/ D 1, Hx is a bijectionof R on R, where R WD R[ f1g. From the relation H0

x.t/ D .B.x/C tD.x//�2 > 0it follows that Hx is strictly increasing on R outside the pole t D � B.x/

D.x/ . Therefore,since Hx.�s/ D I�F .x/ and Hx.1/ D I�K .x/ as shown in the preceding paragraph,Hx is a bijection of Œ�s;1� on the interval ŒITF .x/; ITK .x/�.

Now suppose that ˚ 2 P�s . Then I�˚ .z/ D Hz.˚.z// is holomorphic on CnR.Let x < 0. Then ˚.x/ 2 Œ�s;1/ and hence Hx.�.x// 2 ŒITF .x/; ITK .x/�. Hencethe denominator of Hx.˚.x// does not vanish, because otherwise Hx.˚.x// D 1.Therefore, since the functions A;B;C;D are entire and ˚.z/ is holomorphic onCnRC; I�˚ .z/ D Hz.˚.z// is holomorphic on CnRC: Hence supp�˚ RC byProposition A.15. This proves one direction of Theorem 8.24.

To prove the converse direction we assume that � ¤ �K is a solution of theStieltjes moment problem for s. Then, by formula (8.27) in Theorem 8.18,

I�F.x/ � I�.x/ � I�K .x/ for x < 0: (8.37)

Since �K Š �1, by Theorem 7.13 there is a unique ˚ 2 P such that � D �˚ . Wehave to show that ˚ 2 P�s .

First we verify that ˚.z/ ¤ 0 for z 2 CC � fz 2 C W Im z > 0g. Indeed, assumeto the contrary that ˚.z0/ D 0 for some z0 2 CC. Then, since ˚ 2 P�s , ˚.CC/ isnot open. Therefore ˚ is constant, so that ˚ D 0 2 R. Lemma 6.24 implies that

8.5 Nevanlinna Parametrization for the Indeterminate Stieltjes Moment Problem 195

hT�10 e0; e0i D 0: From the functional calculus of self-adjoint operators we obtain

0 D s0 hT�10 e0; e0i D s0

ZR

y�1dhET0 . y/e0; e0i DZ 1

0

y�1d�˚. y/ <1:

Hence �˚ D 0; which is a contradiction, since �˚ solves an indeterminate Stieltjesmoment problem. This proves that ˚.z/ ¤ 0 for z 2 CC.

Hence � WD 1=˚ is a holomorphic function on CC and we have

�.z/ D 1

˚.z/D 1

H�1z .I�.z//

D �C.z/C I�.z/D.z/

A.z/C I�.z/B.z/: (8.38)

Assume that A.x0/ C I�.x0/B.x0/ D 0 for some x0 < 0. Then, since thenumerator and denominator of the fractional linear transformation do not vanishsimultaneously, C.x0/C I�.x0/D.x0/ ¤ 0 and hence

˚.x0/ WD limz2CC;z!x0

˚.z/ D � limz2CC;z!x0

A.z/C I�.z/B.z/

C.z/C I�.z/D.z/D 0: (8.39)

Since ˚.z/ D H�1z .I�.z//, this implies that ˚.x0/ D H�1

x0 .I�.x0//. By (8.37) wehave H�1

x0 .I�.x0// 2 Œ�s;1�. Since ˚.x0/ D 0 and �s > 0, this is a contradiction.Thus we have proved that A.x/C I�.x/B.x/ ¤ 0 for all x < 0.

Therefore, since � solves the Stieltjes moment problem and hence I� iscontinuous on CnRC, it follows from (8.38) that � has a continuous extensionto CC [ .�1; 0/ with real values on .�1; 0/. Hence, by Schwarz’ reflectionprinciple, � has a holomorphic extension to CnRC: From (8.37) we conclude that˚.x/ D H�1

x .I�.x// 2 Œ�s;1� and therefore �.x/ 2 Œ0; ��1s � for x < 0.

Next we show that�.x/ ¤ 0 on .�1; 0/. Assume to the contrary that�.x0/ D 0for some x0 < 0. Since �.z/ D 1=˚.z/ ¤ 0 on CnRC and �.x/ 2 Œ0; ��1

s � for allx < 0, we have .�1; 0/ \ �.CnRC/ D ;. Hence �.CnRC/ is not open, so � isconstant. Since �.x0/ D 0, �.z/ � 0. But � D 1=˚ with ˚ 2 P, so we have acontradiction.

Putting the preceding together we have shown that ˚ D 1=� has a holomorphicextension to CnRC and ˚.x/ 2 Œ�s;1/ for x 2 .�1; 0/. That is, ˚ 2 P�s . Thisproves the converse direction and completes the proof of Theorem 8.24.

Finally, we prove the two inequalities (8.28) and (8.29) from Theorem 8.18.Since �.x/ ¤ 0 as shown in the paragraph before last, ˚.x/ ¤ 1 and therefore

I�.x/ D Hx.˚.x// ¤ Hx.1/ D I�K .x/. Since I�.x/ � I�K .x/ by (8.37), this yieldsI�.x/ < I�K .x/. This proves the first inequality (8.28).

Now we turn to the proof of (8.29). Let � ¤ �F be a solution of the Stieltjesmoment problem for s. Then � D �˚ , where ˚.z/ WD H�1

z .I�.z// for z 2 CC.First we assume to the contrary that I�.x/ D I�F.x/ for some x < 0. Then we

have˚.x/ D H�1x .I�.x// D H�1

x .I�F .x// D �s, so �.x/ D ��1s is the right end point

of the interval Œ0; ��1s �. Arguing as above, �.CnRC/ is not open, so � and hence˚


are constant. Thus ˚.z/ D ˚.x/ D �s and therefore I�.z/ D Hz.˚.z// D Hz.�s/ DI�F.z/ for all z 2 CnRC. Hence � D �F; which is the desired contradiction.

Now let x D 0. Since� solves the Stieltjes moment problem,˚ 2 P�s , so˚ is ofthe form (8.33). This implies that ˚.z/ 2 Œ�s;1/ and ˚.z/ is monotone increasingon the interval .�1; 0/. Hence the limit �0 WD limz!�0 ˚.z/ 2 Œ�s;C1� exists.We have �0 ¤ �s. (Indeed, otherwise ˚.z/ � �s and the mononiticity of ˚.z/ on.�1; 0/ imply that ˛ D �s and � D 0; then ˚ D �s and hence � D �˚ D �F;

which is a contradiction.) Since A.0/ D D.0/ D 0 and C.0/ D �B.0/ D 1, we haveH0.t/ D t for t 2 R. Clearly, I�.z/ �

R10

y�1d� for z < 0. Then, by the preceding,

Z 1

0

y�1d�F. y/ D I�F .0/ D H0.�s/ D �s < �0

D limz!�0Hz.˚.z// D lim

z!�0 I�.z/ �Z 1

0

y�1d�. y/:

This proves (8.29) for x D 0 and it completes the proofs of (8.28) and (8.29). utWe briefly repeat some facts from the preceding proof. In the parametrization

(8.36) the Friedrichs solution �F corresponds to the Friedrichs parameter ˚ D �s,while the Krein solution �K is obtained for ˚ D 1. The von Neumann solutions ofthe Stieltjes moment problem for s (that is, the solutions � for which CŒx� is densein L2.R; �/) are precisely the measures �t for constants ˚ D t with t 2 Œ�s;C1�:Remark 8.25 The parametrization (7.18) is related to our Nevanlinna parametriza-tion (7.16) by taking �˚�1 instead of ˚ . Hence in this parametrization (7.18)Friedrichs and Krein solutions correspond to the parameters ��1

s and 0, respec-tively. ı

8.6 Weyl Circles for the Indeterminate StieltjesMoment Problem

In this section we suppose that s is an indeterminate Stieltjes moment sequenceand z 2 CC: For t 2 R; let W.z; t/ denote the cone in the complex plane given by

W.z; t/ WD fw 2 C W 0 � arg.w� t/ � � � arg.z/g:

One easily verifies that

w 2 W.z; t/ ” Im.w/ � 0 and Im .z.w � t// � 0: (8.40)

Note that W.z; t/ D tCW.z; 0/ and W.z; t/ W.z; t0/ if t0 � t.

8.6 Weyl Circles for the Indeterminate Stieltjes Moment Problem 197

Proposition 8.26

(i) If ˚ 2 P�s , then ˚.z/ 2 W.z; �s/.(ii) For each w 2 W.z; �s/ there exists a ˚ 2 P�s such that w D ˚.z/:

(iii) Suppose that w 2 W.z; �s/ and Im.z.w � �s// D 0. Then there is a uniquefunction ˚ 2 P�s such that ˚.z/ D w. This function is

˚.z0/ D �s � z.�s � w/

z0 ; z0 2 CC:

Proof

(i) By Proposition 8.23, ˚ 2 P�s is of the form (8.33) with ˛ � �s. Since

Im

�zZ 1

0

d�.x/

x � z

�DZ 1

0

Im

�z

x � z

�d�.x/ � 0

and Im� R1

0.x � z/�1d�.x/

� � 0, (8.40) implies thatR10.x � z/�1d�.x/ 2

W.z; 0/. Therefore, since ˛ � �s, we deduce that

˚.z/ D ˛ CZ 1

0

d�.x/

x � z2 ˛ CW.z; 0/ D W.z; ˛/ W.z; �s/:

(ii) Let z D a C ib and w D u C iv 2 W.z; �s/; where a; b; u; v 2 R: Sincez 2 CC, b > 0. Further, since w 2 W.z; �s/, it follows from (8.40) that v � 0and

Im.z.w � �s// D Im..aC ib/.u � �s C iv// D b.u � �s/C av � 0:(8.41)

First let v D 0. Then u � �s: Hence ˚ WD u 2 P�s and ˚.z/ D w. Nowsuppose v > 0. Then x0 WD v�1.b.u � �s/ C av/ � 0 by (8.41). Definea measure � D v

b jx0 � zj2ıx0 : By Proposition 8.23, ˚.z0/ WD �s CR10 .x �

z0/�1d�.x/ 2 P�s . We compute

˚.z/ D �s C v

bjx0 � zj2.x0 � z/�1 D �s C v

b.x0 � aC ib/ D uC iv D w:

(iii) Suppose that ˚ 2 P�s and ˚.z/ D w: By Proposition 8.23, ˚ is of the form(8.33). Inserting ˚.z/ D w therein we derive

0 D Im.z.w � �s// D Im.z.˛ � �s//CZ 1

0

Im

�z

x � z

�d�.x/

D Im.z/.˛ � �s/C Im.z/Z 1

0

x

jx � zj2 d�.x/:


Fig. 8.1 The transformation Hz and the set KStz

Since Im.z/ > 0 and ˛ � �s, we deduce that ˛ D �s and � D cı0; c � 0. Butthen w D ˚.z/ D �s�cz�1, so that c D z.�s�w/. Hence ˚.z0/ D �s� z.�s�w/

z0 :

Let ˚ be this function. Clearly, ˚.z/ D w. Since Im.z.w � �s// D 0, wederive from (8.41) that c D z.�s � w/ D v.a2b�1 C b/ � 0: This implies that˚ 2 P�s . ut

By Theorem 8.24, the values of Stieltjes transforms of all solutions of theStieltjes moment problem for s are I�� .z/ D Hz.�.z//; where � 2 P�s : FromProposition 8.26 we deduce that these are precisely the numbers of the set

KStz D Hz.W.z; �s/ [ f1g/

which is the right gray shaded area with boundary in Fig. 8.1.Let us consider the boundary of the set KSt

z . Recall that Kz denotes the Weyl circledefined in Sect. 7.3. Then the boundary @KSt

z is the disjoint union of the sets

@�KStz WD @KSt

z \ @Kz D Hz.Œ�s;C1/ [ f1g/;@CKSt

z WD @KStz n@Kz D fHz.w/ W w 2 @W.z; �s//; Im.w/ > 0g:

From the discussion after Theorem 7.13 we know that for each w 2 @�KStz @Kz

there exists a unique solution � of the Hamburger moment problem for s satisfyingI�.z/ D w. Since w 2 @KSt

z , � is also the unique Stieltjes solution such that I�.z/ Dw. Further, � is a von Neumann solution, that is, ord.�/ D 0.

Now let w 2 @CKStz . One easily verifies (see (8.40)) that Im.z.w � �s// D 0.

Therefore, by Proposition 8.26(iii), there is a unique function ˚ 2 P�s such that˚.z/ D w. Hence it follows from Theorem 8.24 that there exists a unique solution�of the Stieltjes moment problem for s satisfying I�.z/ D w. In this case, ord.�/ D 1:

8.8 Notes 199

8.7 Exercises

1. Show s D .sj/j2N0 is a Stieltjes moment sequence if and only if the sequence.s0; 0; s1; 0; s2; : : : / is positive semidefinite.

2. Let s D .sj/j2N0 be a Hamburger moment sequence. Show that .s2j/j2N0 is aStieltjes moment sequence.

3. Let s D .sj/j2N0 be a Stieltjes moment sequence. Show that s2mCn � sksl for allm; n; k; l 2 N0 such that kC l D 2.mC n/.

4. Let s be a Stieltjes moment sequence. Show that ms D inff� W � 2 supp�Fg.5. (A determinate sequence t growing faster than an indeterminate sequence s.)

Find Stieltjes moment sequences s D .sn/n2N0 and t D .tn/n2N0 such that s isStieltjes indeterminate, t is Hamburger determinate, and limn!1 sn

tnD 0:

Hint: Use Corollary 8.22 and Examples 4.18 for ˛ D 14

and 4.23.6. Let .˛n/n2N0 and .ˇn/n2N0 be positive sequences. Prove that there exists a

Stieltjes moment sequence s D .sn/n2N0 such that sn � ˛n and snC1 � ˇnsnfor n 2 N0.

7. Let z 2 CC; t 2 R, and w 2 C.

a. Show that w 2 W.z; t/ if and only if Im.w/ � 0 and Re.z/Im.z/ Im.w/CRe.w/ � t.

b. Show that if w is an interior point of W.z; t/, then Re.z/Im.z/ <

Re.w�t/Im.w�t/ .

8. Suppose that z 2 CC and u 2 CC. Let n 2 N and y; Qy 2 R; y ¤ 0; Qy ¤ 0:a. Show that there are numbers cj > 0 and pairwise distinct points xj 2 R such

that cj.xj � z/ D 1n .u � 2yj/ for j D 1; : : : ; n.

b. Define ˚y.z0/ WD �y.nC 1/CPnjD1

cjxj�z0 ; z

0 2 CC. Show that ˚y 2 P is a

rational function of degree n satisfying ˚y.z/ D u.c. Show that ˚y ¤ ˚Qy for y ¤ Qy.

9. Suppose that s is an indeterminate Stieltjes moment sequence and z 2 CC. Let vbe an interior point of W.z; �s/, w an interior point of KSt

z , and n 2 N.

a. Show that there are infinitely many rational functions ˚ 2 P�s of degree nsuch that ˚.z/ D v.

b. Show that there are infinitely many solutions � of the Stieltjes momentproblem for s such that ord.�/ D n and I�.z/ D w.

Hint for b: Use the construction sketched in Exercise 8. Show that the numbersxj can be chosen positive for small y > 0; details can be found in [Ge].

8.8 Notes

The theory of the Stieltjes moment problem goes back to T. Stieltjes’ famousmemoir [Stj], which contains many basic results.


The Friedrichs parameter has been identified in [Pd1]. Since these papers usethe parametrization in the form (7.18), their number is our ��1

s . The Nevanlinnaparametrization for indeterminate Stieljtes moment sequences was first given byH.L. Pedersen [Pd1]; it is also contained in [Sim1].

The use of the operators TF and TK and the corresponding approximants is dueto B. Simon [Sim1]. Our operator-theoretic approach is based on [Sim1] and [Ge].

Part IIThe One-Dimensional Truncated

Moment Problem

Chapter 9The One-Dimensional Truncated Hamburgerand Stieltjes Moment Problems

In this chapter we are concerned with the following problem:Let s D .sj/mjD0 be a real m-sequence, where m 2 N0. When does there exist a

Radon measure � on R such that sj DRRxj d�.x/ for all j D 0; : : : ;m?

This is the truncated Hamburger moment problem and in the affirmative case s iscalled a truncated Hamburger moment sequence. If we require in addition that themeasure � is supported on RC, we get the truncated Stieltjes moment problem.

In Sects. 9.1 and 9.3 the special case of a positive definite 2n-sequence isstudied in detail. Using quasi-orthogonal polynomials Gauss’ quadrature formulasare derived (Theorems 9.4 and 9.6) and a one-parameter family of .n C 1/-atomicsolutions is constructed (Theorem 9.7). The associated reproducing kernel spaceand the Christoffel function are investigated in Sect. 9.3. In the short Sect. 9.2 weapply some result from Sect. 9.1 to reprove Hamburger’s and Markov’s theorem inthe positive definite case.

The remaining part of the chapter deals with positive semidefinite finitesequences. In Sect. 9.4 such sequences are characterized by integral representations(Theorems 9.15 and 9.19). In Sect. 9.5 the Hankel rank of a positive semidefinite 2n-sequence is introduced. The integral representation and the Hankel rank enter intothe treatment of truncated moment problems in Sect. 9.6. Here basic existencetheorems for the truncated Hamburger and Stieltjes moment problems in theeven case m D 2n (Theorems 9.27 and 9.36) and in the odd case m D 2n C 1

(Theorems 9.32 and 9.35) are obtained. Further, neccesary and sufficient conditionsfor the uniqueness of the representing measures are given.

Let us recall some standard notations. The real polynomials of degree at most nare denoted by RŒx�n. For a sequence s D .sj/mjD0 and 2n � m, the Hankel matrixHn.s/ is defined by Hn.s/ D .siCj/

ni;jD0, the corresponding Hankel determinant is

Dn.s/ WD detHn.s/, and Ls is the Riesz functional on RŒx�m defined by Ls.xj/ D sj.


203

204 9 The One-Dimensional Truncated Hamburger and Stieltjes Moment Problems

9.1 Quadrature Formulas and the Truncated MomentProblem for Positive Definite 2n-Sequences

Throughout this section, we assume that n 2 N and s D .sj/2njD0 is a real positivedefinite 2n-sequence. That s is positive definite means that

nXk;lD0

skCl�k�l > 0 for all .�0; : : : ; �n/T 2 RnC1; .�0; : : : ; �n/T ¤ 0: (9.1)

In terms of the Riesz functional Ls on RŒx�2n, condition (9.1) is equivalent to therequirement Ls. p2/ > 0 for all p 2 RŒx�n, p ¤ 0.

Lemma 9.1 Let s2jC1, j � n, be given real numbers. There exist real numbers s2ifor i 2 N, i � nC 1; such that s D .sk/k2N0 is a positive definite sequence.

Proof Since .sj/2njD0 is positive definite, the Hankel matrix Hn.s/ is positive definite.Hence the Hankel determinant Dn.s/ of s is positive. Let s2nC2 be a real number.The Hankel determinant DnC1 for the sequence .sj/

2nC2jD0 is of the form

DnC1 D s2nC2Dn.s/C p.s0; : : : ; s2n; s2nC1/:

Here p is a real polynomial p in s0; : : : ; s2n; s2nC1 that does not depend on s2nC2.Since Dn.s/ > 0, we have DnC1 > 0 for sufficiently large s2nC2. Proceeding byinduction this extension procedure leads to a positive definite sequence s. ut

Combining Lemma 9.1 with Hamburger’s Theorem 3.8 we obtain the following

Corollary 9.2 The truncated Hamburger moment problem for each positive definitereal 2n-sequence is solvable.

As in Chap. 5 we define a scalar product h�; �is on RŒx�nC1 by

h p; qis D Ls. pq/; p; q 2 RŒx�nC1: (9.2)

For this scalar product on RŒx�nC1 and at a few other places we need an extension ofs to a positive definite .2nC 2/-sequence. Such an extension exists by Lemma 9.1.We fix this extension. Note that (9.2) also depends on the numbers s2nC1; s2nC2:

Definition 9.3 A polynomial P 2 RŒx�nC1;P ¤ 0, is called quasi-orthogonal ofrank nC 1 if

Ls.Pxj/ D 0 for j D 0; : : : ; n � 1: (9.3)

(Since deg.Pxj/ � 2n, (9.3) does not require an extension of the 2n-sequence s.)

Proceeding as in Sect. 5.1 we obtain a unique sequence p0; : : : ; pnC1 of orthonor-mal polynomials of the unitary space .RnC1Œx�; h�; �is/ such that degpk D k and the

9.1 Quadrature Formulas and the Truncated Moment Problem for Positive. . . 205

leading coefficient of pk is positive for k D 0; : : : ; nC 1. Since then Ls. pkxj/ D 0

for j < k, pn and pnC1 are quasi-orthogonal of rank nC 1. Therefore, for any t 2 R,

P.t/.x/ D pnC1.x/C tpn.x/ (9.4)

is quasi-orthogonal of rank n C 1 and degree n C 1. Up to constant multiples allquasi-orthogonal polynomials of rank nC 1 and degree nC 1 are of this form.

In what follows we fix a quasi-orthogonal polynomial P of rank nC1 and degreenC 1. By Proposition 5.28, P has nC 1 real zeros �1 < �2 < � � � < �nC1. Let usintroduce the quantities

�j.x/ WD P.x/

.x � �j/P0.�j/and mj WD Ls.�

2j / > 0 ; j D 1; : : : ; nC 1; (9.5)

Q.z/ WD Ls;x

�P.x/� P.z/

x � z

�(9.6)

and define an .nC 1/-atomic measure �P by

�P DnC1XjD1

mjı�j : (9.7)

Since P.x/�P.z/x�z is a polynomial in x of degree at most n, the functional Ls;x applies

to this polynomial and Q.z/ is a polynomial of degree at most n. Clearly,

�j.x/ D .x � �1/ : : : .x � �j�1/.x � �jC1/ : : : .x � �nC1/.�j � �1/ : : : .�j � �j�1/.�j � �jC1/ : : : .�j � �nC1/

;

�j.�i/ D ıij for i; j D 1; : : : ; nC 1: (9.8)

Now can can state and prove the first main theorem of this section.

Theorem 9.4 For each polynomial f 2 RŒx�2n we have

Ls. f / DnC1XjD1

mjf .�j/ �Z

f .x/ d�P.x/: (9.9)

Further, for j D 1; : : : ; nC 1 and z 2 Cnf�1; : : : ; ; �nC1g,

mk D Q.�k/

P0.�k/D Ls.�k/ and � Q.z/

P.z/D

nC1XjD1

mj

�j � z�Z

d�P.x/

x � z: (9.10)


Proof For each p 2 RŒx�n we have the Lagrange interpolation formula

p.x/ DnC1XjD1

�j.x/ p.�j/: (9.11)

(To prove (9.11) it suffices to note the the difference of both sides is a polynomialof degree at most n that vanishes at nC 1 points �j, so it is identically zero.)

Now fix f 2 RŒx�2n. Since deg.P/ D nC 1, there are polynomials qf 2 RŒx�n�1and pf 2 RŒx�n such that

f .x/ D P.x/qf .x/C pf .x/: (9.12)

The defining relation (9.3) for the quasi-orthogonal polynomial P yields Ls.Pqf / D0, so that Ls. f / D Ls. pf /. Further, since P.�j/ D 0 and hence pf .�j/ D f .�j/ by(9.12), combining the relations Ls. f / D Ls. pf / and (9.11) we obtain

Ls. f / DnC1XjD1

Ls.�j/f .�j/: (9.13)

Applying this formula to f D �2k by using (9.5) and �j.�k/ D ıjk by (9.8) we get

mk D Ls.�2k / D

nC1XjD1

Ls.�j/�k.�j/2 D Ls.�k/ ; k D 1; : : : ; nC 1: (9.14)

Inserting (9.14) into (9.13) we obtain (9.9).Combining (9.14) with (9.5) and (9.6) we get

mk D Ls.�k/ D Ls

�P.x/

.x � �k/P0.�k/

�D 1

P0.�k/Ls

�P.x/� P.�k/

x � �k�D Q.�k/

P0.�k/:

This yields the first half of (9.10).Finally, we prove the second half of (9.10). Since degP D n C 1, P has n C 1

simple real zeros, and degQ � n, there exists a partial fraction decomposition

� Q.z/

P.z/D

nC1XjD1

�j

�j � z(9.15)

with �j 2 R. Here the coefficients �j are

�j D limz!�j

.z � �j/Q.z/P.z/

D Q.�j/

P0.�j/D mj

9.1 Quadrature Formulas and the Truncated Moment Problem for Positive. . . 207

by the first part of (9.10). Inserting �j D mj into (9.15) gives the second half of(9.10). ut

Formula (9.9) is usually called a Gaussian quadrature formula. If the functionalLs is of the form Ls. f / D

Rf .x/d�.x/ for some measure �, then (9.9) reads as

Zf .x/ d�.x/ D

nC1XjD1

mjf .�j/: (9.16)

The numbers �j are called nodes and the numbers mj weights of the quadratureformula. For general functions f the sum on the right of (9.16) can be considered asan approximation of the integral on the left.

Theorem 9.4 says that the identity (9.16) holds for all polynomials f 2 RŒx�2n.But (9.16) does not hold for f 2 RŒx�2nC2. For instance, if f .x/ D QnC1

jD1 .x � �j/2,then the right-hand side of (9.16) vanishes, but Ls. f / D

Rf d� > 0, because s is

positive definite.Before we continue we derive a nice application of formula (9.9).

Corollary 9.5 Let P and QP be two quasi-orthogonal polynomials of rank nC 1 anddegree nC 1 with zeros �1 < � � � < �nC1 and Q�1 < � � � < Q�nC1, respectively. Theneither QP is a multiple of P or the sequences .�1; � � � ; �nC1/ and . Q�1; � � � ; Q�nC1/are stricly interlacing, that is, for any k there is a j such that �k < Q�j < �kC1. Inparticular, if t; Qt 2 R and t ¤ Qt, then the zeros of the quasi-orthogonal polynomialsP.t/ and P.Qt/ defined by (9.4) are strictly interlacing.

Proof Fix k 2 f1; : : : ; ng and put

fk.x/ D 1

.x � �k/.x � �kC1/nC1YjD1.x � �j/2:

Then fk 2 RŒx�2n, so (9.9) applies and yields Ls. fk/ D 0, since fk.�j/ D 0 forj D 1; : : : ; nC 1. Applying the same formula with P replaced by QP we get

Ls. fk/ DnC1XjD1

emjfk. Q�j/ D 0: (9.17)

The zeros of fk are precisely the numbers�l and we have fk.x/ < 0 for x 2 .�k; �kC1/and fk.x/ � 0 for x � �k and x � �kC1. Therefore, if there were no j such that�k < Q�j < �kC1, then all (!) numbers Q�j would have to be contained in the zero setof fk. This implies that �k D Q�k for all k, so P is a constant multiple of QP. ut

As in Sect. 5.2 the orthonormal polynomials pk; k � n; satisfy the recurrencerelation (5.9); let an be the corresponding Jacobi coefficient from (5.9).

In the case P D pnC1 we have the following stronger result than Theorem 9.4.


Theorem 9.6 If P D pnC1, then formula (9.9) holds for all f 2 RŒx�2nC1 and

m�1j D anpn.�j/p

0nC1.�j/ D

nXkD0

pk.�j/2; j D 1; : : : ; nC 1: (9.18)

Conversely, if there are real numbers Q�j and positive numbersemj, j D 1 : : : ; nC1,such that Q�1 < � � � < Q�nC1 and

Ls. f / DnC1XkD1

emjf . Q�k/ for all f 2 RŒx�2nC1; (9.19)

then Q�j D �j and emj D mj for jD1; : : : ; n C 1 and formula (9.19) coincides with(9.9).

Proof Let f 2 RŒx�2nC1. Then the polynomial qf in (9.12) belongs to RŒx�n.Therefore, since P D pnC1, we have Ls.Pqf / D 0. Proceeding now as in the proofof Theorem 9.4 it follows that formula (9.9) holds for f as well.

Next we prove (9.18). Since p0; : : : ; pnC1 are orthonormal polynomials for theextended sequence (by Lemma 9.1), the formulas (5.47) and (5.51) proved inSect. 5.5 are valid. We set z D �j in the Christoffel formula (5.47) and rememberthat pnC1.�j/ D 0. Inserting the definition (9.5) of �j.x/ for P D pnC1 we derive

nXkD0

pk.x/pk.�j/ D anpnC1.x/pn.�j/

x � �j D an�j.x/p0nC1.�j/pn.�j/:

Applying the functional Ls by using that p0 D s�1=20 and mj D Ls.�j/ we get

1 D s�10 Ls.1/ D Ls. p0/p0.�j/ D

nXkD0

Ls. pk/pk.�j/ D anpn.�j/p0nC1.�j/mj

which yields the first equality of (9.18). The second equality follows from the firstby setting x D �j in formula (5.51) and using once again that pnC1.�j/ D 0.

Finally, we prove the uniqueness assertion. We set

QP.x/ D .x � Q�1/ : : : .x � Q�nC1/

and define Q�j andemj by (9.8) with P replaced by QP and �j by Q�j.By deg Q�j D n and P D pnC1, we have Ls.P Q�j/ D 0. Since deg.P Q�j/ � 2nC 1,

formula (9.19) applies to P Q�j as well. Using the relation Q�j. Q�k/ D ıkj (by 9.8))we obtain 0 D Ls.P Q�j/ D emjP. Q�j/: Thus, P. Q�j/ D 0, since emj > 0: That is, thenumbers Q�j are the zeros of P. Therefore, Q�j D �j for j D 1; : : : ; n C 1 and QP is a

9.2 Hamburger’s Theorem and Markov’s Theorem Revisited 209

constant multiple of P D pnC1. Hence Q�j D �j, so that mj D Ls.�j/ D Ls. Q�j/ D emj

by (9.10) for j D 1; : : : ; nC 1. utThe next theorem summarizes and restates some of the preceding results on the

quadrature formula (9.9) in terms of the truncated Hamburger moment problem.

Theorem 9.7 Suppose that s D .sj/2njD0 is a real positive definite 2n-sequence. Let�t denote the .nC1/-atomic measure �P.t/ for the quasi-orthogonal polynomial P.t/defined by (9.4) and (9.7), respectively.

(i) Each measure �t , t 2 R, is a solution of the truncated Hamburger momentproblem for s, that is, we have sj D

Rxj d�t.x/ for j D 0; : : : ; 2n:

(ii) For t; Qt 2 R, t ¤ Qt, the atoms of �t and �Qt are strictly interlacing, so �t and�Qt have disjoint supports.

(iii) For any � 2 R that is not a zero of pn.x/, there exists a unique t 2 R such that� is an atom of �t.

(iv) For each real number s2nC1 there is a unique .nC 1/-atomic measure � suchthat s0; : : : ; s2nC1 are the moment of �. In fact, � is the measure �0 when thesequence .sj/

2nC1jD0 is extended to a positive definite .2nC2/-sequence .sj/2nC2

jD0 .

Proof (i) and (ii) follow from formula (9.9) and Corollary 9.5, respectively.

(iii) Setting t WD �pn.�/�1pnC1.�/, the number � is a zero of the quasi-orthogonalpolynomial

P.t/.x/ D pnC1.x/ � pn.�/�1pnC1.�/pn.x/

and hence an atom of �t. (ii) yields the uniqueness assertion.(iv) We extend .sj/

2nC1jD0 to a positive definite real sequence .sn/n2N0 (by Lemma 9.1)

and put P WD P.0/ � pnC1. Then, by Theorem 9.6, (9.9) holds for thepolynomials x0; : : : ; x2nC1 which means that the measure �0 has the momentss0; : : : ; s2nC1. The uniqueness of � D �0 follows from the second part ofTheorem 9.6. ut

We restate assertion (iv) of the preceding theorem separately as

Corollary 9.8 For each positive definite real sequence .sj/2njD0 and each realnumber � there exists a unique .nC 1/-atomic measure �� such that

sj DZ

xj d��.x/ for j D 0; : : : ; 2n and � DZ

x2nC1 d��.x/:

9.2 Hamburger’s Theorem and Markov’s Theorem Revisited

In this section, we assume that s D .sn/n2N0 is a real positive definite sequence.Recall that pn and qn are the orthogonal polynomials of the first and the second kind,respectively, associated with s.


We apply Theorem 9.4 with P D pnC1; k D n C 1; and set �k WD �P. Thenqk D Q by (9.6). Let �

.n/1 < � � � < �

.n/n denote the zeros of pk and m.k/j the

corresponding weights. From (9.9), applied with f D xj, and (9.10) we obtain

sj DZR

xj d�k.x/ ; j D 0; : : : ; k � 1; (9.20)

�qk.z/pk.z/

DkX

jD1

m.k/j

�.k/j � z

DZR

d�k.x/

x � z; z 2 Cnf�.k/1 ; : : : ; ; �.k/k g: (9.21)

Since �k.R/ D s0 for all k, .s�10 �k/k2N0 is a sequence of probability measures.

Hence it follows from Theorem A.9 that there exists a subsequence .�kn/n2N0 whichconverges vaguely to some measure � 2 MC.R/.

Let L�k be the linear functional on RŒx�k�1 defined by L�k . f / D Rf d�k. Then,

by (9.20), .L�k /k2N is a directed sequence of linear functionals on RŒx� according toDefinition 1.17. Hence limn!1 L�kn D L� by Lemma 1.18 (or Theorem 1.20) and

L�.xj/ D L�kn .xj/ DZR

xj d�kn D sj for kn > j

by (9.20). Therefore, � is a representing measure for s, that is, � solves theHamburger moment problem for s. Thus we have given a third proof of the mainimplication (ii)!(i) of Hamburger’s theorem 3.8 for positive definite sequences.

By formula (5.60) the sequence .�.k/1 /k2N is decreasing and the sequence

.�.k/k /k2N is increasing. Therefore, the limits

˛s WD limk!1�

.k/1 2 f�1g [R and ˇs WD lim

k!1�.k/k 2 R [ fC1g

exist, see Corollary 5.33. Let Js denote the smallest closed interval which contains.˛s; ˇs/. That is, Js is the smallest closed interval which contains all zeros of theorthonormal polynomials pk; k 2 N. Note that ˛s, ˇs, and Js depend only on thesequence s, but not on any representing measure. Since supp�k Œ�.k/1 ; �.k/k � Js

for k 2 N by (9.21), it follows that supp� Js: In the literature [Chi1] a solutionof the moment problem with support contained in Js is called a natural solution.

Now we use formula (9.21) to develop a second approach to Markov’s theo-rem 6.29.

Theorem 9.9 Suppose that s is a determinate positive definite sequence s. If � andJs are as above, then

� limk!1

qk.z/

pk.z/DZJs

d�.x/

x � zfor z 2 CnJs; (9.22)

where the convergence is uniform on compact subsets of CnJs.

9.3 The Reproducing Kernel and the Christoffel Function 211

Proof By Theorem 1.20, .�k/k2N converges vaguely to �. Let z 2 CnJs. Then thefunction fz.x/ D 1

x�z is in C0.Js/. Therefore, since supp�k Js and supp� Js,Proposition A.9 implies that

limk!1

ZJs

d�n.x/

x � zDZJs

d�.x/

x � z: (9.23)

Then (9.22) is obtained by inserting (9.21) on the left-hand side of (9.23).Now we prove the assertion concerning the uniform convergence. Let K be a

compact subset of CnJs. Then c WD dist .K;Js/ > 0. By (9.21), the convergence in(9.22) is uniform on K provided this holds for the convergence in (9.23). Let " > 0be given. Since K is compact, there are finitely many numbers z1; : : : ; zk such thatthe open discs centered at zj with radius " cover K. Given z 2 K we choose j suchthat jz � zjj < ". Therefore, we have

ˇˇ 1

x � z� 1

x � zj

ˇˇ D jz � zjjj.x � z/.x � zj/j �

"

c2for x 2 Js:

Let k 2 N. Then, using that supp�k Js; supp� Js, and �k.R/ D �.R/ D s0,we derive

ˇˇZ

d�k.x/

x � z�Z

d�.x/

x � z

ˇˇ (9.24)

�ˇˇ Z d�k.x/

x � z�Z

d�k.x/

x � zj

ˇˇCˇˇ Z d�k.x/

x � zj�Z

d�.x/

x � zj

ˇˇC

ˇˇ Z d�.x/

x � zj�Z

d�.x/

x � z

ˇˇ

� s0"

c2Cˇˇ Z d�.x/

x � zj�Z

d�k.x/

x � zj

ˇˇC s0

"

c2: (9.25)

For each j D 1; : : : ; k, the middle term in (9.25) converges to zero by (9.23). Hencefor sufficiently large k the expression in (9.24) is less than 3"s0c�2. This proves thatthe convergence in (9.23) is uniform on K. ut

9.3 The Reproducing Kernel and the Christoffel Function

Throughout this section, s D .sj/2njD0 is again a real positive definite 2n-sequence.Let .CŒx�n; h�; �is/ be the complex Hilbert space with scalar product defined by

h p; qis D Ls. p q/, p; q 2 CŒx�n, where q.x/ WDPj ajxj for q.x/ DPj ajx

j 2 CŒx�n:As above, p0; : : : ; pn denote the corresponding orthonormal polynomials. Recall

that pj 2 RŒx� and hence pj.y/ D pj.y/ for y 2 C. Then

Kn.x; y/ WDnX

kD0pk.x/pk.y/ ; x; y 2 C; (9.26)


is a reproducing kernel on the Hilbert space .CŒx�n; h�; �is/, that is, we have

p.y/ D h p.x/;Kn.x; y/is and hKn.x; y/; p.x/is D p.y/; p 2 CŒx�n; y 2 C:

(9.27)

As a sample we prove the second equality; the proof of the first equality is similar.We write p DPn

jD0 cjpj.x/ and derive

hKn.x; y/; p.x/is;x DnX

j;kD0pk.y/ cj h pk.x/; pj.x/is D

nXj;kD0

cjpk.y/ ıjk D p.y/:

Note that Kn.z; z/ DPnkD0 jpk.z/j2 > 0 for all z 2 C, since p0.z/ > 0.

Definition 9.10 The n-th Christoffel function is

n.z/ WD Kn.z; z/�1 D

� nXkD0jpk.z/j2

��1; z 2 C: (9.28)

The following proposition expresses the kernel Kn.x; y/ and the Christoffelfunction n.z/ D Kn.z; z/�1 in terms of determinants involving the moments.

Proposition 9.11 For x; y 2 C and n 2 N0 we have

Kn.x; y/ D �D�1n

ˇˇˇˇˇˇˇ

0 1 x : : : xn

1 s0 s1 : : : sny s1 s2 : : : snC1y2 s2 s3 : : : snC2: : : : : : : : : : : : : : :

yn sn snC1 : : : s2n

ˇˇˇˇˇˇˇ

: (9.29)

Proof Let G.x; y/ denote the determinant on the right. Fix y 2 C and j 2 f0; : : : ; ng.We compute Ls;x.xjG.x; y//. That is, we multiply the first row in the determinant byxj and then apply the functional Ls;x by using the multilinearity of determinants. As aresult, the first row of G.x; y/ is replaced by .0; sj; sjC1; : : : ; snCj/. Then we subtractthe . jC 2/-th row .yj; sj; sjC1; : : : ; snCj/. This does not change the determinant andyields the first row .�yj; 0; : : : ; 0/. Finally, we develop the resulting determinantafter the first row and get Ls;x.G.x; y/xj/ D �yjDn:

Set H.x; y/ WD �D�1n G.x; y/. Since G.x; y/ 2 RŒx; y�, we have

hxj;H.x; y/is;x D �D�1n Ls;x.x

j G.x; y// D �D�1n .�yjDn/ D yj:

This implies that h p.x/;H.x; y/is;x D p.y/ for all p 2 CŒx�n. In particular,

hKn.x; z/;H.x; y/is;x D Kn.y; z/ for y; z 2 C: (9.30)

9.3 The Reproducing Kernel and the Christoffel Function 213

On the other hand, it follows from the definition of G that G.z; y/ D G.y; z/.This implies that H.z; y/ D H.y; z/. Therefore, by the second equality of (9.27),

hKn.x; z/;H.x; y/is;x D H.z; y/ D H.y; z/ for y; z 2 C: (9.31)

Comparing (9.30) and (9.31) we get Kn.y; z/ D H.y; z/, which is the assertion. utThe next proposition characterizes the function n by an extremal property.

Proposition 9.12 For any z 2 C we have

min fLs. p p/ W p 2 CŒx�n; p.z/ D 1g D Kn.z; z/�1 D n.z/ (9.32)

and the minimum is attained at p if and only if p.x/ D Kn.x; z/n.z/.

The proof will be derived below from Proposition 9.14. Since the latter result isused in Sect. 10.7 in the real setting, we formulate it in the real and complex cases.It is based on the following fact from elementary Hilbert space theory.

Lemma 9.13 Let f ¤ 0 be an element of a real or complex unitary spaceH. Then

min fkgk2 W g 2 H; hg; f i D 1g D k fk�2 (9.33)

and the minimum (9.33) is attained at g if and only if g D fk fk�2.Proof Put g0 D fk fk�2. If hg; f i D 1, then hg � g0; f i D hg; f i � k fk�2h f ; f i D 0:Thus g � g0?g0. Therefore, by Pythagoras’ theorem,

kgk2 D kg0 C g � g0k2 D kg0k2 C kg � g0k2:

Hence the minimum is attained at g if and only if g D g0. utProposition 9.14 Let K D C or K D R. Suppose that s D .sj/2njD0 is apositive definite real sequence and let p0; : : : ; pn be the corresponding orthonormalpolynomials. Then, for any z 2 K we have

min fLs. p p/ W p 2 KŒx�n; p.z/ D 1g D� nX

jD0jpj.z/j2

��1(9.34)

and the minimum is attained at p if and only if

p.x/ DnX

jD0pj.x/ pj.z/

� nXiD0jpi.z/j2

��1: (9.35)


Proof Let p 2 KŒx�n. We develop p with respect to the orthonormal basis f pjg ofthe unitary space .KŒx�n; h�; �is/ and obtain

p DnX

jD0cjpj; where cj D h p; pjis; and Ls. p p/ D kpk2s D

nXjD0jcjj2:

Then p.z/ D Pj cjpj.z/. Hence, the problem in (9.34) is to minimize

Pj jcjj2 for

g WD .c0; : : : ; cn/T 2 KnC1 under the constraintP

j cjpj.z/ D 1. This is solved by

Lemma 9.13, applied to the unitary space KnC1 and f WD . p0.z/; : : : ; pn.z/ /T . Theminimum is k fk�2 and it is attained if and only if cj D pj.z/ k fk�2 for j D 0; : : : ; n.Inserting these facts yields the asserted formulas (9.34) and (9.35). utProof of Proposition 9.12 We apply Proposition 9.14 in the case K D C. By (9.28)the minimum in (9.34) is n.z/. Comparing (9.35) and (9.26) we conclude that theminimum is attained if and only if p.x/ D Kn.x; z/n.z/. ut

9.4 Positive Semidefinite 2n-Sequences

In the rest of this chapter we investigate positive semidefinite finite sequences.The following theorem characterizes the positive semidefiniteness of a 2n-

sequence in terms of an integral representation (9.36).

Theorem 9.15 For any real sequence s D .sj/2njD0 the following five statements areequivalent:

(i) s is positive semidefinite, that is,

nXk;lD0

skClckcl � 0 for .c0; : : : ; cn/T 2 RnC1:

(ii) Hn.s/ � 0.(iii) Ls. p2/ � 0 for all p 2 RŒx�n.(iv) There exists a Radon measure � on R and a real number a � 0 such that

RŒx�2n L1.R; �/,

sj DZR

xj d�.x/ for j D 0; : : : ; 2n � 1; and s2n D aCZR

x2n d�.x/:

(9.36)

(v) There exists a k-atomic measure � on R, where k � 2n C 1, and a constanta � 0 such that (9.36) holds.

9.4 Positive Semidefinite 2n-Sequences 215

Proof The equivalence of (i), (ii), and (iii) is easily verified by the same reasoningused in the proof of Hamburger’ theorem 3.8; we do not repeat it here.

(iii)!(v) The proof is based on Proposition 1.26. Let X WD R[ f1g denote theone point compactification of R. Then the functions

uj.x/ WD xj.1C x2n/�1; j D 0; : : : ; 2n;

extend to continuous functions on X by setting uj.1/ D 0 for j D 0; : : : ; 2n � 1and u2n.1/ D 1. Let E be the span of these functions in C.X IR/.

We define a linear functional QL on E by QL.uj/ D Ls.xj/; j D 0; : : : ; 2n. Letf 2 EC. Then g WD .1 C x2n/f 2 RŒx�2n is nonnegative on R. Therefore, byProposition 3.1, we have g D p2 C q2 with polynomials p; q 2 RŒx�n. Thus weobtain f D . p2 C q2/.1C x2n/�1 and hence QL. f / D Ls. p2 C q2/ by the definitionof QL. Since Ls. p2/ � 0 and Ls.q2/ � 0 by (iii), QL. f / � 0. Thus, Proposition 1.26applies with g D u2n and there is a k-atomic positive measure Q� on X , k � 2nC 1;such that QL. f / D R f d Q� for f 2 E.

Set a D Q�.f1g/. We define atomic measures O� and � on R by O�.M/ WD Q�.M/for M R and d� WD .1C x2n/�1d O�. For j D 0; : : : ; 2n we compute

sj D Ls.xj/ D QL.uj/ D

ZXuj.x/ d Q� D auj.1/

CZR

xj

1C x2nd O� D aıj;2n C

ZR

xj d�;

which proves (v).(v)!(iv) is trivial.(iv)!(ii) Let .c0; : : : ; cn/T 2 RnC1. Using (9.36) we derive

nXk;lD0

skClckcl D ac2n CnX

k;lD0

ZR

ckclxkCld� D ac2n C

ZR

� nXkD0

ckxk

�2d� � 0;

(9.37)

since a � 0. This proves (ii). utThe positive semidefiniteness of s is not sufficient for representing all numbers sj

as moments sj DRxjd�, see Example 9.17 below. It implies only a representation

(9.36) with s2n �Rx2n d�. Some authors consider (9.36) as the “right version”

of the Hamburger truncated moment problem. We require that s2n is the .2n/-thmoment of � as well, that is, s2n D

Rx2n d�. To obtain the latter equality

additional conditions are needed; they depend on whether or not we are dealingwith the Hamburger or Stieltjes truncated moment problem and in the even or oddcase.


Our main existence results on the Hamburger truncated moment problem arederived from the following corollary.

Corollary 9.16 Let s D .sj/2njD0 be a real sequence and suppose that Hn.s/ � 0.(i) There is a k-atomic positive measure � on R, k � 2nC 1; such that

Ls. p/ DZR

p.x/ d� for p 2 RŒx�2n�1: (9.38)

If Ls. f 2/ D 0 for some f 2 RŒx�n, then supp� Z. f /.(ii) Suppose that there exists a polynomial f 2 RŒx� of degree n such that Ls. f 2/ D

0. Then s is a truncated Hamburger moment sequence and supp� Z. f / foreach representing measure � of s:

Proof

(ii) Let f .x/ DPnkD0 cnxn. Using (9.36) and (9.37) we get

0 D Ls. f2/ D

nXk;lD0

skClckcl D ac2n CZR

f .x/2 d�: (9.39)

Since deg. f / D n, cn ¤ 0. Hence a D 0, so that s is truncated moment sequenceby (9.36). Inserting a D 0 into (9.39), Proposition 1.23 implies that supp� Z. f /.

(i) The first assertion is only a reformulation of Theorem 9.15 (i)!(v).If deg. f / D n, the second assertion follows from (ii). If deg. f / < n,then Ls. f 2/ D

Rf 2 d� D 0 by (9.38), so that supp� Z. f / again by

Proposition 1.23. utExample 9.17 Let s D .0; 0; 1/. Then H1.s/ � 0. Since s0 D 0, s cannot begiven by a Radon measure on R, but s has a representation (9.36) with � D 0

and a D 1. ıExample 9.18 Let s D .16; 0; 4; 0; 4/. Then the sequence s has a representation(9.36) with � D 8ı� 1

2C 8ı 1

2and a D 3. Since detH2.s/ > 0, s is positive definite,

so it also has representing measures on R by Corollary 9.2; one such measure is� WD 2ı�1 C 12ı0 C 2ı1. ı

The next theorem is the counterpart of Theorem 9.15 for the positive half-line.

Theorem 9.19 Let s D .sj/mjD0 be a real sequence and set Es WD .s1; : : : ; sm/: Thefollowing are equivalent:

(i) m D 2n: Hn.s/ � 0 and Hn�1.Es/ � 0, that is,nX

k;lD0skClckcl � 0 and

n�1Xk;lD0

skClC1ckcl � 0 for .c0; : : : ; cn/T 2 RnC1:

9.5 The Hankel Rank of a Positive Semidefinite 2n-Sequence 217

m D 2nC 1: Hn.s/ � 0 and Hn.Es/ � 0, that is,nX

k;lD0skClckcl � 0 and

nXk;lD0

skClC1ckcl � 0 for .c0; : : : ; cn/T 2 RnC1:

(ii) m D 2n W Ls. p2/ � 0 and Ls.xq2/ � 0 for p 2 RŒx�n; q 2 RŒx�n�1.m D 2nC 1 W Ls. p2/ � 0 and Ls.xq2/ � 0 for p; q 2 RŒx�n:

(iii) There exists a Radon measure � on RC and a real number a � 0 such thatRŒx�m L1.RC; �/,

sj DZ 1

0

xj d�.x/ for j D 0; : : : ;m� 1; and sm D aCZ 1

0

xm d�.x/:

(9.40)

(iv) There exists a k-atomic measure � on RC, where k � mC 1, and a constanta � 0 such that (9.40) holds.

Proof The proof follows the same reasoning as the proof of Theorem 9.15. Wesketch only the proof of the main implication (ii)!(iv). Let X D RC [ fC1gdenote the one point compactification of RC and E the span of continuous functions

uj.x/ WD xj.1C xm/�1; j D 0; : : : ;m;

on X , where uj.C1/ WD 0 if j D 0; : : : ;m� 1 and um.C1/ WD 1.We define a linear functional QL on E by QL.uj/ WD Ls.xj/. Let f 2 EC. Then

g D .1 C xm/f 2 RŒx�m is nonnegative on RC. Therefore, by Proposition 3.2, thepolynomial g is of the form g D pC xq, where p 2P2

n; q 2P2

n�1 if m D 2n andp; q 2 P2

n if m D 2nC 1. Because of these formulas the conditions in (ii) implythat QL. f / � 0. Therefore, by Proposition 1.26, QL is given by some k-atomic positivemeasure Q� on X , where k � mC 1. Continuing as in the proof of Theorem 9.15 theformulas in (9.40) are derived. ut

9.5 The Hankel Rank of a Positive Semidefinite 2n-Sequence

Let s D .sj/2njD0 be a real sequence. We define the kernel vector space of s by

Ns WD f p 2 RŒx�n W Ls. pq/ D 0 for q 2 RŒx�n g:

The first assertion of the following lemma gives another description of Ns.


Lemma 9.20

(i) For p D PnjD0 ajxj 2 RŒx�n, set Ep WD .a0; : : : ; an/T 2 RnC1. Then the map

p 7! Ep is a bijection of Ns on kerHn.s/. In particular,

dim Ns D dim kerHn.s/ and rankHn.s/ D nC 1 � dimNs: (9.41)

(ii) If s is positive semidefinite, then Ns D fp 2 RŒx�n W Ls. p2/ D 0 g:Proof

(i) For p DPnjD0 ajxj 2 RŒx�n and q DPn

jD0 bjxj 2 RŒx�n, we compute

L. pq/ D2nX

i;jD0Ls.x

iCj/aibj D2nX

i;jD0siCjaibj D EqT Hn.s/Ep:

Therefore, p 2 Ns if and only if Ep 2 kerHn.s/: Hence p 7! Ep is a bijection ofNs on kerHn.s/. Obviously, this implies (9.41).

(ii) The left-hand side is contained in the right-hand side by setting p D q.Since s is positive semidefinite, Ls. f 2/ � 0 for f 2 RŒx�n. Hence the

Cauchy–Schwarz inequality (2.7) holds, that is, Ls. pq/2 � Ls. p2/Ls.q2/ forp; q 2 RŒx�n. Therefore, Ls. p2/ D 0 implies Ls. pq/ D 0. ut

Clearly, Ns is a linear subspace of RŒx�n. Therefore, if Ns ¤ f0g, there exists aunique monic polynomial f 2 RŒx�n, fs ¤ 0; of lowest degree in Ns.

Definition 9.21 If Ns ¤ f0g, we call fs the minimal polynomial associated with sand the number rk.s/ D deg. fs/ the Hankel rank of s. In the case Ns D f0g we setrk .s/ D nC 1.

The Hankel rank plays a crucial role in the truncated moment problem.Clearly, rk .s/ D 0 if and only if fs D 1, or equivalently, s0 D � � � D sn D 0:

Further, by definition and Lemma 9.20(i), we have rk .s/ D n C 1 if and only ifNs D f0g; or equivalently, rankHn.s/ D nC 1.

Let us write the Hankel matrix Hn.s/ as

Hn.s/ D Œv0; : : : ; vn�;

where vj D .sj; : : : ; sjCn/T , j D 0; : : : ; n; are the column vectors of Hn.s/.

The second assertion of the following lemma is usually called Frobenius’ lemmain the literature, see e.g. [Gn, Lemma X.10.1].

Lemma 9.22 Suppose that 1 � rk .s/ � n:

(i) rk .s/ is the smallest integer r, 1 � r � n; such that the column vector vr is inthe span of v0; : : : ; vr�1, or equivalently, there are reals �0; : : : ; �r�1 such that

sjCr D �r�1sjCr�1 C � � � C �0sj for j D 0; : : : ; n: (9.42)

9.5 The Hankel Rank of a Positive Semidefinite 2n-Sequence 219

If this is true, then we have

fs.x/ D xr � �r�1xr�1 � � � � � �1x � �0:

(ii) Drk.s/�1.s/ ¤ 0 and Drk.s/.s/ D 0:Proof

(i) Since 1 � rk .s/ � n, the minimal polynomial fs is defined and we havedeg. fs/ � 1; as noted above. Let f .x/ D xr � �r�1xr�1 � � � � � �1x � �0;where r 2 f1; : : : ; ng and �0; : : : ; �r�1 2 R: Then

Ls.xjf / D sjCr � �r�1sjCr�1 � � � � � �1sjC1 � �0sj; j D 0; : : : ; n: (9.43)

Hence f 2 Ns if and only if (9.42) holds, or equivalently, vr D �0v0 C � � � C�r�1vr�1. Thus the smallest r such that vr 2 spanfv0; : : : ; vr�1g is obtained ifand only if the monic polynomial f has the lowest possible degree in Ns; thatis, if f D fs.

(ii) Let Ak D Œv0; : : : ; vk�1� denote the matrix with columns v0; : : : ; vk�1. By(i), rk .s/ is the smallest index r such that vr belongs to the linear span ofv0; : : : ; vr�1.

Since vr is in the span of v0; : : : ; vr�1, we have rankArC1 < r C 1: ThusDr.s/ D 0:

Because r is the smallest such number, v0 : : : ; vr�1 are linearly independentand hence rankAr D r: Further, by (i) there are real numbers �0; : : : ; �r�1 suchthat the equations (9.42) hold. It follows from (9.42) that each row of Ar is alinear combination of the r preceding rows. Therefore, each row of Ar is in thespan of the first r rows of Ar. Since rankAr D r is also the maximal number oflinearly independent rows, the first r rows of Ar are linearly independent. Thisimplies Dr�1.s/ ¤ 0. ut

Example 9.23 Let n D 4 and s D .1; 1; 1; 1; 0; 0; 0; 0; 0/. Since D4.s/ D 0; weeasily derive from Proposition 9.25(i) that rk .s/ D 4. Then fs D x4; D0.s/ DD3.s/ D 1; and D1.s/ D D2.s/ D 0: Since Ls..1 � x2/2/ D �1, s is not positivesemidefinite! ı

In the rest of this section, the sequence s is positive semidefinite. Then we haveHn.s/ � 0 and Ls. p2/ � 0 for p 2 RŒx�n. From formula (9.41) and some elementarylinear algebra we obtain the following lemma; we omit its simple proof.

Lemma 9.24 If s is positive semidefinite, then the following are equivalent:

(i) s is positive definite.(ii) Hn.s/ � 0.

(iii) Dn.s/ ¤ 0.(iv) rk.s/ D nC 1.

Some important properties of rk .s/ are collected in the next proposition.


Proposition 9.25 Suppose that sD.sj/2njD0 is positive semidefinite and rk.s/ � n.

(i) If n � 1 � rk .s/ � 0, then pfs 2 Ns for p 2 RŒx�n�1�rk .s/:

(ii) rk .s/ � rankHn.s/ � rk .s/C 1.(iii) rk .s/ D rankHn.s/ if and only if xn�rk .s/fs 2 Ns.(iv) rk .s/ is the smallest number r 2 N0 such that Dr.s/ D 0:More precisely,

D0.s/ > 0; : : : ;Drk .s/�1.s/ > 0; Drk.s/.s/ D 0; : : : ;Dn.s/ D 0: (9.44)

(In the case rk .s/ D 0, the set of inequalities should be omitted.)Proof

(i) Since fs 2 Ns, Corollary 9.16(i) applies, so there exists a Radon measure �such that supp� Z. fs/ and Ls.q/ D

Rq d� for q 2 RŒx�2n�1. Therefore, if

p 2 RŒx�n�1�rk .s/, then . pfs/2 2 RŒx�2n�2 and pfs 2 Ns. since

Ls.. pfs/2/ D

ZZ. fs/

p.x/2fs.x/2 d�.x/ D 0:

(ii) First we note that the following set is a basis of the vector space RŒx�n W

fxi; xjfs W i; j 2 N0; 0 � i � rk.s/ � 1; 0 � j � n � rk.s/g: (9.45)

If rk.s/ D n, then the assertion is obvious. Hence we can assume thatrk.s/ < n: By (i) we have xjfs 2 Ns for j D 0; : : : ; n � 1 � rk.s/. Thus,dimNs � n � rk.s/.

Now we prove that dim Ns � n C 1 � rk.s/. Assume the contrary, thatis, dimNs � n C 2 � rk.s/. Since (9.45) is a vector space basis of RŒx�n, itfollows from (i) that there exists a polynomial f 2 RŒx�n; f ¤ 0;with deg. f / <rk.s/ D deg. fs/ in Ns. This contradicts the choice of fs. Hence dim Ns �nC 1 � rk.s/: Therefore, rk .s/ � rankHn.s/ � rk .s/C 1 by (9.41).

(iii) By (9.41), rk .s/ D rankHn.s/ if and only if dimNs D nC 1 � rk.s/. Sincerk.s/ is the lowest degree of nonzero polynomials in Ns and (9.45) is a basisof RŒx�n, we conclude from (i) that the latter is equivalent to xn�rk .s/fs 2 Ns.

(iv) Set r WD rk .s/: If r D 0, then s0 D � � � D sn D 0, as noted above. Hence sj D0 for all j, since c is positive semidefinite, so that Dk.s/ D 0 for k D 0; : : : ; n.

Assume now that r � 1: Then, by Lemma 9.22(ii), Dr�1.s/ ¤ 0 andDr.s/ D 0: Since Hn.s/ � 0 by assumption, Dr�1.s/ > 0. Hence the matrixHr�1.s/ is positive definite and therefore Dk.s/ > 0 for k D 0; : : : ; r� 1. SinceDr.s/ D 0; the sequence .sj/2rjD0 is not positive definite, hence neither is thesequence .sj/2kjD0 and so Dk.s/ D 0 for k D r C 1; : : : ; n: This completes theproof of (9.44). ut

9.6 Truncated Hamburger and Stieltjes Moment Sequences 221

Let us illustrate the preceding in an important special case.

Example 9.26 Suppose s has a k-atomic representing measure� DPkjD1mjıtj with

k � n. Let f 2 RŒx�n. Then f 2 Ns if and only if Ls. f 2/ DPkjD1mjf .tj/2 D 0, or

equivalently, f .t1/ D � � � D f .tk/ D 0. Therefore, fs.x/ D .x � t1/ : : : .x � tk/ andrk.s/ D k: By Theorem 9.27 (i)!(iii) below, we also have rankHn.s/ D k. ı

9.6 Truncated Hamburger and Stieltjes Moment Sequences

In this section, we settle the truncated Hamburger and Stieltjes moment problemsfor real sequences s D .sj/2njD0 (even case) and s D .sj/2nC1

jD0 (odd case).

Theorem 9.27 (The Hamburger Truncated Moment Problem; Even Case) Forany real sequence s D .sj/2njD0 the following are equivalent:

(i) s is a truncated Hamburger moment sequence.(ii) There exist reals s2nC1; s2nC2 such that HnC1.Qs/ � 0, where Qs WD .sj/2nC2

jD0 .(iii) Hn.s/ � 0 and rankHn.s/ D rk.s/:

If rk.k/ � n, or equivalently, if Ns ¤ f0g, these conditions are equivalent to:(iv) Hn.s/ � 0 and xn�rk.s/fs 2 Ns:

(v) Hn.s/ � 0 and there exists a polynomial of degree n in Ns:

Suppose that (i) is satisfied. Then s has an rk.s/-atomic representing measure.Further, s has a unique representing measure � if and only if rk.s/ � n, orequivalently, Dn.s/ D 0. The atoms of this measure are the zeros of fs and

jsupp�j D jZ. fs/j D rk.s/ D rankHn.s/: (9.46)

Proof(i)!(ii) From (i) and Corollary 1.25 it follows that s has a finitely atomic

representing measure �. Hence � has all moments. Then, setting sl DRxl d� for

l D 2nC 1; 2nC 2, Qs is a truncated moment sequence and hence HnC1.Qs/ � 0.(ii)!(i) Since HnC1.Qs/ � 0, Corollary 9.16(i) applies with s replaced by Qs and n bynC 1. Then Eq. (9.38) therein implies that sj D

Rxj d� for j D 0; : : : ; 2n.

Next we treat the case rk.s/ D nC1: Then Ns D f0g and rankHn.s/ D nC1 by(9.41) and Definition 9.21. The implication (i)!(iii) is clear. We prove (iii)!(i).Since Hn.s/ � 0 and rankHn.s/ D n C 1, the sequence s is positive definite.Therefore, by Theorem 9.7, s has a one-parameter family of .n C 1/-atomicrepresenting measures. Thus in the case rk.s/ D nC1 all assertions of the theoremare proved.

In the rest of this proof we assume that rk.s/ � n: Then Ns ¤ f0g and theminimal polynomial fs is defined.


(i)!(iv) By (i), s has a representing measure �. Hence Hn.s/ � 0. Since fs 2 Ns,we have supp� Z. fs/ by Proposition 1.23 and therefore

Ls..xn�rk.s/fs/

2/ DZZ. fs/

x2.n�rk.s//fs.x/2 d� D 0:

That is, xn�rk.s/fs 2 Ns and (iv) is proved.(iv)!(v) is trivial, since xn�rk.s/fs belongs to Ns and has degree n.(v)!(i) follows at once from Corollary 9.16(ii).(iii)$(iv) is clear by Proposition 9.25(iii).This completes the proof of the equivalence of statements (i)–(v).Suppose that (i) holds. Let � be a representing measure for s. As noted above,

supp� Z. fs/, so � is k-atomic with k � deg. fs/ D rk.s/ D rankHn.s/ by(iii). By assumption, rk.s/ � n. Thus, k � n: Hence, as noted in Example 9.26,k D rk.s/: The preceding proves all equalities of (9.46).

Let Q� be another representing measure for s. Let Z. fs/ D ft1; : : : ; tkg. Thenthere are nonnegative numbers mj; nj such that � DPk

jD1mjıtj and Q� DPkjD1 njıtj .

Since k � n, there are interpolation polynomials pi 2 RŒx�n such that pi.tj/ D ıijfor i; j D 1; : : : ; n. Then Ls. pi/ D

Rpi d� D mi and Ls. pi/ D

Rpi d Q� D ni, so

that mi D ni for all i. Hence � D Q�. Thus s has a unique representing measure ifrk.s/ � n: utRemark 9.28 The proof of (i)!(iv) uses the same reasoning as the proof ofProposition 9.25(i), now applied with p D xn�rk.s/. Such arguments and theuniqueness proof based on interpolation polynomials often appear (for instance, inthe proofs of Theorem 10.7 and Proposition 17.18) and in slightly different settingsin this book. ı

We illustrate Theorem 9.27 with two simple examples.

Example 9.29 Let n D 2 and s D .sj/4jD0, where s0 D s1 D s2 D s3 D 1, s4 D ˛ �1: Clearly, s has a representation (9.36) with � D ı1 and a D ˛ � 1 � 0. Hence

H2.s/ D0@1 1 11 1 1

1 1 ˛

1A � 0 :

Case 1: ˛ > 1.Then fs D x � 1, rk.s/ D 1, and Ns D R�fs. Since x2�1fs … Ns, s is not a truncatedHamburger moment sequence. Note that .x � 1/ 2 Ns, but .x � 1/2 … Ns:

Case 2: ˛ D 1.Then fs D x � 1 and rk.s/ D 1, but xfs 2 Ns, so s is a truncated moment sequencewith unique representing measure � D ı1: ı


Example 9.30 Set � WD ı0C 1k2nık for fixed k 2 N and n � 2. Clearly, the moment

sequence s D .sj/2njD0 of � is given by

s0 D 1C k�2n; sj D k�.2n�j/ for 1 � j < 2n; s2n D 1:

In the limit k ! 1 we obtain the positive semidefinite sequence Qs D.1; 0; : : : ; 0; 1/: It is not a moment sequence, since fQs D x 2 N .Qs/ andxn D xn�1fQs … N .Qs/. This shows that the set of all truncated moment 2n-sequencesis not closed in R2nC1. ı

The following lemma about positive block matrices is a special case of Theo-rem A.24. We identify Rk with the column matrices Mk;1.R/.

Lemma 9.31 Let k 2 N, A 2 Mk.R/, b 2 Rk, c 2 R, and suppose that A � 0:

Consider the block matrix QA 2 MkC1.R/ defined by

QA D�A bbT c

�: (9.47)

(i) QA � 0 if and only if b D Au for some u 2 Rk and c � uTAu.(ii) Let b D Au with u 2 Rk and c � uTAu. Then rank QA D rankA if and only if

c D uTAu.

The three remaining existence theorems of this section use Lemma 9.31.

Theorem 9.32 (The Hamburger Truncated Moment Problem; Odd Case) Fora real sequence s D .sj/2nC1

jD0 the following three statements are equivalent:

(i) s is a truncated Hamburger moment sequence.(ii) There exists a real number s2nC2 such that HnC1.Qs/ � 0, where Qs WD .sj/2nC2

jD0 .(iii) Hn.s/ � 0 and .snC1; : : : ; s2nC1/T 2 range .Hn.s//:

Suppose that (i) holds and set Os WD .sj/2njD0: Then s has an rk.Os/-atomic representingmeasure. Further, s has a unique representing measure if and only if rk.Os/ � n, orequivalently, Dn.Os/ D 0I this unique measure is rk.Os/-atomic and its set of atomsis the zero set Z. fOs/ of fOs.

Proof The proof of (i)$(ii) is almost verbatim the same as for Theorem 9.27. In theproof of (ii)!(i) we apply Corollary 9.16 with n replaced by nC1. Hence Eq. (9.38)holds for j D 0; : : : ; 2nC 1 which gives (i).

To prove (ii)$(iii) we consider HnC1.Qs/ as a block matrix (9.47) with

A D Hn.s/; b D .snC1; : : : ; s2nC1/T ; c D s2nC2: (9.48)

(ii)!(iii) Since HnC1.Qs/ � 0, we have Hn.s/ � 0 and b 2 range .Hn.s// by theonly if direction of Lemma 9.31(i).

(iii)!(ii) Then b 2 range .Hn.s//, say b D Hn.s/u with u 2 RnC1. Therefore,choosing s2nC2 � uTHn.s/u, we have HnC1.Qs/ � 0 by the if part of Lemma 9.31(i).


This completes the proof of the equivalences (i)–(iii). Now suppose that (i) holds.First we assume that rk.Os/ D n C 1. Then rankHn.Os/ D n C 1 by Proposi-

tion 9.25(ii). Hence, since Hn.s/ D Hn.Os/, there exists a v 2 RnC1 such that b DHn.s/v. We fix v and choose s2nC2 > vTHn.s/v. Then, applying Lemma 9.31(ii) tothe block matrix HnC1.Qs/ yields rankHnC1.Qs/ > rankHn.s/ D rankHn.Os/ D nC 1.Therefore, rankHnC1.Qs/ D n C 2. Hence, by Corollary 9.2 (or by Theorem 9.27(iii)!(i)), Qs is a truncated moment sequence. Obviously, any representing measurefor Qs is one for s. Since these measures have the moment s2nC2, they are differentas s2nC2 varies. Thus s has a one-parameter family of representing measures.

Assume now that rk.Os/ � n. Any representing measure � for s is a representingmeasure for Os. Therefore, by Theorem 9.27 applied to Os, � is uniquely determinedand has the properties stated above. ut

The next proposition shows how Lemma 9.31 can be used to obtain additionalinformation about the structure of the extended sequence Qs in Theorem 9.27(ii).

Proposition 9.33 Let s D .sj/2njD0 be a positive semidefinite real sequence suchthat r WD rk.s/ � n: Suppose that there exist real numbers s2nC1; s2nC2 such thatHnC1.Qs/ � 0, where Qs WD .s0; : : : ; s2n; s2nC1; s2nC2/. Further, let �0; : : : ; �r�1 be thenumbers from Eq. (9.42). Then

sjCr D �0sj C � � � C �r�1sjCr�1 for j D 0; : : : ; 2nC 1 � r; (9.49)

s2nC2 � �0s2nC2�r C � � � C �r�1s2nC1: (9.50)

There is equality in (9.50) if and only if rk.s/ � rankHn.s/ D rankHnC1.Qs/:Proof We consider the Hankel matrix HnC1.Qs/ as a block matrix (9.47) with entries(9.48). Since HnC1.Qs/ � 0, we have Hn.s/ � 0 and Lemma 9.31(i) applies.Therefore, we have b D Au with u 2 RnC1 and c � uTAu. Equation b D Aumeans that

snCkC1 DnX

lD0skClul; k D 0; : : : ; n: (9.51)

We prove by induction that Eq. (9.49) holds for j D 0; : : : ; 2n C 1 � r: If j D0; : : : ; n, then (9.49) is just one of the equations (9.42). Assume that (9.49) is truefor all j, where n � j � 2n� r. Using (9.51) and the induction hypothesis (9.49) wederive

sjC1Cr DnX

lD0sjCr�nClul D

nXlD0

r�1XiD0

sj�nClCi�iul

Dr�1XiD0

� nXlD0

sj�nClCiul

��i D

r�1XiD0

sjC1Ci�i;


which is (9.49) for j C 1. This completes the induction proof of (9.49). (Thepreceding computation is not valid for j D 2n C 1�r, since (9.51) does not holdfor k D nC 1.)

By Lemma 9.31(i), s2nC2 D c � uTAu: Using (9.51) and (9.49) we compute

ssnC2 � utAu D utb DnX

lD0snC1Clul D

nXlD0

r�1XiD0

snC1�rClCi�iul Dr�1XiD0

s2nC2�rCi�i;

which is the inequality (9.50).Theorem 9.27 (ii)!(iii) implies that rk.s/ D rankHn.s/. According to

Lemma 9.31(ii), we have rankHn.s/ D rankHnC1.Qs/ if and only if s2nC2 �c D uTAu, or equivalently, if there is equality in (9.50). utRemark 9.34

1. Let us retain the assumptions of Proposition 9.33. Equation (9.49) is a recursiverelation for the numbers sj; j � 2n C 1: In particular, s2nC1 is determined by(9.49) for j D 2nC 1� r. From Lemma 9.31(ii) it follows that each real numberssnC2 satisfying (9.50) gives an extension Qs such that HnC1.Qs/ � 0: In particular,we can choose s2nC2 WD �0s2nC2�rC� � �C�r�1s2nC1; then we have rankHn.s/ DrankHnC1.Qs/.

2. Let s D .sj/2njD0 be a positive definite real sequence. Then rankHn.s/ Dn C 1: (This case is not covered by Proposition 9.33, since the assumptionsof Proposition 9.33 and Theorem 9.27 imply that s is a moment sequence andrankHn.s/ D rk.s/ � n.) Since Hn.s/ is regular, the vector b from (9.48) is inthe range of A D Hn.s/, say b D Hn.s/u. Hence, by Lemma 9.31(i), for arbitraryreals s2nC1; s2nC2 such that s2nC2 � uTHn.s/u, the extension Qs D .sj/

2nC2jD0 of s

satisfies HnC1.Qs/ � 0; so s is a moment sequence by Theorem 9.27. If we chooses2nC2 > uTHn.s/u, Lemma 9.31(ii) implies that rankHn.Qs/ D n C 2; so Qs isalso positive definite. This recovers the extension procedure for positive definitesequences from Lemma 9.1. ıNow we turn to the truncated Stieltjes moment problem and begin with the odd

case. Let Es D .s1; : : : ; sm/ denote the shifted sequence of s D .s0; s1; : : : ; sm/.Theorem 9.35 (The Stieltjes Truncated Moment Problem; Odd Case) A realsequence s D .sj/2nC1

jD0 is a truncated Stieltjes moment sequence if and only if

Hn.s/ � 0; Hn.Es/ � 0; and .snC1; : : : ; s2nC1/T 2 range .Hn.s//: (9.52)

Suppose s is a truncated Stieltjes moment sequence. Then s has an rk.Os/-atomicrepresenting measure on RC. Moreover, the sequence s has a unique representingmeasure onRC if and only if rk.s/ � n or rk.Es/ � n, or equivalently, Dn.s/ D 0or Dn.Es/ D 0.Proof By Theorem 9.32 (i)$(ii), s is a truncated Hamburger moment sequence ifand only if Hn.s/ � 0 and .snC1; : : : ; s2nC1/T 2 range .Hn.s//:


Suppose s has a representing measure � supported on RC. Then the positive (!)measure � given by d� D xd� has the moment sequence Es, so that Hn.Es/ � 0.

Conversely, assume that Hn.Es/ � 0. By Theorem 9.32, s has an rk.Os/-atomicrepresenting measure � D Prk.Os/

jD1 mjıtj . Since rk.Os/ � n C 1; there are Lagrangeinterpolation polynomials pi 2 RŒx�n such that pi.tj/ D ıij, i; jD1; : : : ;rk.Os/. Then

L.xp2i / DZ

xpi.x/2 d�.x/ D

rk.Os/XjD0

mjtjp2i .tj/ D miti � 0:

Since � is rk.Os/-atomic, we have mi > 0 and therefore ti � 0. Thus � is supportedon RC and s is a truncated Stieltjes moment sequence.

We turn to the proof of the uniqueness assertions. Note that rk.s/ � n (resp.rk.Es/ � n) if and only if Dn.s/ D 0 (resp. Dn.Es/ D 0) by Lemma 9.24 (iii)$(v).

First, let us assume that

Dn.s/ ¤ 0 and Dn.Es/ ¤ 0: (9.53)

Let Qs D .sj/2nC3jD0 with unknown s2nC2; s2nC3. There are polynomials f ; g such that

DnC1.Qs/ D s2nC2Dn.s/C f .s0; : : : ; s2nC1/; (9.54)

DnC1.EQs/ D s2nC3Dn.Es/C g.s1; : : : ; s2nC2/: (9.55)

Since Hn.s/ � 0 and Hn.Es/ � 0, both determinants in (9.53) are positive. Hencewe can find s2nC2 > 0 such that (9.54) is positive and there exists a c > 0 such that(9.55) is positive for s2nC3 � c. Then the sequence Qs fulfills (9.52) with n replacedby n C 1. (Since DnC1.Qs/ > 0, HnC1.Qs/ is positive definite and hence its range isRnC2.) Therefore, by the existence assertion of Theorem 9.35 proved above, Qs is atruncated Stieltjes moment sequence. Since s2nC3 � c was arbitrary, we obtain afamily of different measures on RC which have the same moments s0; : : : ; s2nC1:

Now suppose that Dn.s/ D 0. Then, by Theorem 9.32, s has a uniquerepresenting measure on R, so we have uniqueness on RC as well.

Finally, we suppose that Dn.Es/ D 0: Let �1 and �2 be representing measuresfor s. We will show that �1 D �2: Define measures �j by d�j D xd�j, j D 1; 2.Then Es is a truncated Hamburger sequence with representing measures �j. SinceDn.Es/ D 0; Theorem 9.27 yields �1 D �2. Since d�j D xd�j, this implies that�1.N/ D �2.N/ for all Borel sets N Rnf0g. We have s0 D �j.f0g/C�j.Rnf0g/,j D 1; 2. Therefore, since �1.Rnf0g/ D �2.Rnf0g/, we obtain �1.f0g/ D �2.f0g/.Thus �1 D �2. This completes the proof of the uniqueness assertions. utTheorem 9.36 (The Stieltjes Truncated Moment Problem; Even Case) A realsequence s D .sj/2njD0 is a truncated Stieltjes moment sequence if and only if

Hn.s/ � 0; Hn�1.Es/ � 0; and .snC1; : : : ; s2n/T 2 range .Hn�1.Es//:(9.56)

In this case s has an rk.s/-atomic representing measure on RC.


Further, the truncated Stieltjes moment sequence s has a unique representingmeasure supported onRC if and only if rk.s/ � n, or equivalently, if Dn.s/ D 0:Proof First assume that the three conditions (9.56) are fulfilled. Our aim is to applyTheorem 9.35 to the extended sequence Qs D .sj/2nC1

jD0 for some real number s2nC1:Since b WD .snC1; : : : ; s2n/T 2 range .Hn�1.Es// by (9.56), there exists a u 2

Rn such that b D Hn�1.Es/u. Put c D s2nC1 WD uTHn�1.Es/u and A WD Hn�1.Es/:Since Hn�1.Es/ � 0 by (9.56), the block matrix QA in (9.47) is positive semidefiniteby Lemma 9.31(i). But QA is the Hankel matrix Hn.EQs/, where EQs D .s1; : : : ; s2nC1/.Thus, Hn.EQs/ � 0.

We set w WD .s0; : : : ; sn�1/T and write Hn.s/ as block matrix

Hn.Qs/ D Hn.s/ D

0BBB@

s0 s1 : : : sns1 s2 : : : snC1::::::

: : ::::

sn snC1 : : : s2n

1CCCA D

�w Hn�1.Es/sn bT

�:

Applying Hn.Qs/ to the column vector .0; u/T 2 RnC1 we obtain

Hn.Qs/�0

u

�D�Hn�1.Es/ubTu

�D�buTHn�1.Es/u

�D�bs2nC1

�;

that is, .snC1; : : : ; s2n; s2nC1/T 2 range .Hn.Qs//. Thus, since Hn.Qs/ D Hn.s/ � 0

by (9.56) and Hn.EQs/ � 0 as shown above, Qs satisfies (9.52). Therefore, byTheorem 9.35, Qs is a truncated Stieltjes moment sequence which has an rk.s/-atomic representing measure. Hence this holds for s as well.

Conversely, suppose s is a truncated Stieltjes moment sequence. It is obvious thatHn.s/ � 0 and Hn�1.Es/ � 0, so it remains to show the range condition in (9.56).We argue as in the proofs of Theorems 9.27 and 9.32. By Corollary 1.25, s has afinitely atomic representing measure � with atoms in RC. Setting sl D

Rxl d�.x/

for l D 2nC 1; 2nC 2, Qs D .sj/2nC2jD0 is a truncated Stieltjes moment sequence. Since

the Hankel matrix Hn.EQs/ is a block matrix (9.47) with

A D Hn�1.Es/; b D .snC1; : : : ; s2n/T ; c D s2nC1

and Hn.EQs/ � 0, Lemma 9.31(i) implies that b 2 range .Hn�1.Es//.It remains to verify the uniqueness assertions. First, suppose that Dn.s/ D 0.

Then the representing measure of s for the even truncated Hamburger problem isunique by Theorem 9.27, so it is unique for the even Stieltjes case as well.

Now assume that Dn.s/ ¤ 0. As above we set Qs D .sj/2nC1jD1 with unknown real

s2nC1. We consider Hn.EQs/ as a block matrix

Hn.EQs/ D�Hn�1.s.1// bbT s2nC1

�; where b WD .snC1; : : : ; s2n/T :


Since s is a truncated moment sequence, it has a finitely atomic representing measure�. Putting s2nC1 D

Rx2nC1 d�, Qs is an odd truncated Stieltjes moment sequence,

so Hn.EQs/ � 0 by Theorem 9.35. From Lemma 9.31(i) we obtain Hn.EQs/ � 0

if s2nC1 is sufficiently large. Fix such a number s2nC1. Then Qs satisfies (9.52).Indeed, Hn.Qs/ D Hn.s/ � 0, since s is a moment sequence, and range .Hn.Qs// DRnC1, since Dn.Qs/ D Dn.s/ ¤ 0: Therefore, by Theorem 9.35, Qs is a Stieltjesmoment sequence. The representing measures for Qs are different as s2nC1 varies,but all of them have the same moments s0; : : : ; s2n. Thus, we have nonuniquenessif Dn.s/ ¤ 0. ut

9.7 Exercises

1. Determine Hankel rank and minimal polynomial:

a. s D .3; 0; 2; 0; 2; 0; 2/, m D 6.b. s D .2; 2; 4; 8; 16/, m D 4.c. Set sj D 1 for j D 0; 1; 2; 3 and sj D 0 for j D 4; : : : ; 2n; where n � 4.

2. Decide whether or not the following sequences are truncated Hamburger momentsequences:

a. s D .4; 5; 9/, m D 3.b. s D .2; 0; 2; 0; 5/, m D 4.c. s D .3; 2; 6; 10; 18; 35/, m D 6.d. s D .3; 0; 2; 0; 2; 0; 2; 0/, m D 7.

3. Decide whether or not the following sequences are truncated Stieltjes momentsequences:

a. s D .1; 2; 4; 8; 20/, m D 4.b. s D .3; 5; 9; 17; 33/, m D 4.c. s D .3; 2; 2; 2; 2; 3/, m D 5.d. s D .3; 5; 11; 29/, m D 3.

9.8 Notes

Quadrature formulas of the form (9.9) were discovered by C.F. Gauss (1814) andstudied by K.G.J. Jacobi [Jac] and E.B. Christoffel [Chl]. Our treatment of thepositive definite case in Sect. 9.1 follows N.I. Akhiezer and M.G. Krein [AK, §1].The representation formula (9.36) for positive semidefinite sequences was provedby E. Fischer [Fi]. The results for the truncated Hamburger and Stieltjes momentproblems in Sect. 9.6 are due to R. Curto and L. Fialkow [CF1].

Chapter 10The One-Dimensional Truncated MomentProblem on a Bounded Interval

Throughout this chapter a and b are fixed real numbers such that a < b and m 2 N.We consider the truncated moment problem on the interval Œa; b�:

Given a real m-sequence s D .sj/mjD0, when is there a Radon measure � on Œa; b�

such that sj DR ba xj d�.x/ for j D 0; : : : ;m?

In this case we say that s is a truncated Œa; b�-moment sequence and � is arepresenting measure for s.

In Sect. 10.1 truncated Œa; b�-moment sequences are described in terms ofpositivity conditions (Theorems 10.1 and 10.2). If a solution exists, there are alwaysatomic solutions. The main part of this chapter deals with atomic solutions of “smallsize”. In Sect. 10.2 the cone SmC1 of all moment sequences and the index of atomicrepresenting measures are introduced. Boundary points of SmC1 are characterizedas moment sequences with unique representing measures and of index at most m(Theorem 10.7). The rest of this chapter is devoted to a detailed study of interiorpoints of SmC1. In Sects. 10.3, 10.4, and 10.6 representing measures of indexm C 1 and m C 2 are investigated. Each interior point of SmC1 has precisely tworepresenting measures of index m C 1 (Theorem 10.17) and for each � 2 Œa; b�a distinguished measure with root � and index at most m C 2 (Corollary 10.13).The maximal mass of a point among all representing measures is studied inSect. 10.5 (Theorem 10.21). In Sect. 10.7 orthogonal polynomials are developed anda description of the maximal mass in terms of orthonormal polynomials is given(Theorem 10.29).

10.1 Existence of a Solution

First we collect some notation that will be used throughout this chapter.Let s D .sj/mjD0 be a real sequence. Recall that Ls is the Riesz functional on RŒx�n

given by Ls.xj/ D sj, j D 0; : : : ; n; and Hk.s/, 2k � m, denotes the Hankel matrix


229

230 10 The One-Dimensional Truncated Moment Problem on a Bounded Interval

Hk.s/ WD .siCj/ki; jD0. The shifted sequence Es is Es WD .s1; : : : ; sm/ D .sjC1/m�1

jD0 :Recall that A � 0 means that the matrix A D AT 2 Mm.R/ is positive semidefinite.

The following notation differs between the two cases m D 2n and m D 2nC 1:

H2n.s/ WD Hn.s/ � .siCj/ni; jD0;

H2n.s/ WD Hn�1..b � E/.E � a/s/ � ..aC b/siCjC1 � siCjC2 � absiCj/n�1i; jD0;

H2nC1.s/ WD Hn.Es� as/ � .siCjC1 � asiCj/ni; jD0;

H2nC1.s/ WD Hn.bs� Es/ � .bsiCj � siCjC1/ni; jD0:

Here the lower index always refers to the highest moment in the correspondingmatrix and the highest moment occurs only in the right lower corner. The upper andlower bar notation and the lower indices will be seen to be useful later. An advantageis that they allow us to treat the even and odd cases at once.

Further, we abbreviate the corresponding Hankel determinants by

Dm.s/ WD detHm.s/; Dm.s/ WD detHm.s/: (10.1)

For f D PkjD0 ajxj 2 RŒx�k let Ef WD .a0; : : : ; ak/T 2 RkC1 denote the coefficient

vector of f . Then for p; q 2 RŒx�n and f ; g 2 RŒx�n�1 simple computations yield

Ls. pq/ D EpT H2n.s/Eq; Ls..b � x/.a � x/fg/ D Ef T H2n.s/Eg; (10.2)

Ls..x � a/pq/ D EpT H2nC1.s/Eq; Ls..b � x/pq/ D EpT H2nC1.s/Eq: (10.3)

From Proposition 3.3 we restate the formulas (3.8) and (3.9) describing thepositive polynomials Pos.Œa; b�/m on Œa; b� of degree at most m:

Pos.Œa; b�/2n D˚f C .b � x/.x � a/g W f 2 ˙2

n ; g 2 ˙2n�1�; (10.4)

Pos.Œa; b�/2nC1 D˚.b � x/f C .x � a/g W f ; g 2 ˙2

n

�: (10.5)

The following two existence theorems rely essentially on these descriptions.

Theorem 10.1 (Truncated Œa; b�-Moment Problem; Even Case m D 2n) For areal sequence s D .sj/2njD0 the following statements are equivalent:

(i) s is a truncated Œa; b�-moment sequence.(ii) Ls. p2/ � 0 and Ls..b � x/.x � a/q2/ � 0 for p 2 RŒx�n and q 2 RŒx�n�1.

(iii) H2n.s/ � 0 and H2n.s/ � 0:Theorem 10.2 (Truncated Œa; b�-Moment Problem; Odd Case m D 2nC 1) Fora real sequence s D .sj/2nC1

jD0 the following are equivalent:

(i) s is a truncated Œa; b�-moment sequence.(ii) Ls..x � a/p2/ � 0 and Ls..b � x/p2/ � 0 for all p 2 RŒx�n.

(iii) H2nC1.s/ � 0 and H2nC1.s/ � 0:

10.2 The Moment Cone SmC1 and Its Boundary Points 231

Proofs of Theorems 10.1 and 10.2:(i)$(ii) We apply Proposition 1.9 to the subspace E D RŒx�m of C.Œa; b�IR/.

Then Ls is a truncated Œa; b�-moment functional if and only if Ls. p/ � 0 for allp 2 EC D Pos.Œa; b�/m: By (10.4) and (10.5) this is equivalent to condition (ii).

(ii)$(iii) follows at once from the identities (10.2) and (10.3). utRemark 10.3 Since p2 D .b � a/�1Œ.b � x/p2 C .x � a/p2�, condition (ii) inTheorem 10.2 implies that Ls. p2/ � 0 for p 2 RnŒx�: ı

By Theorem 1.26, we can have finitely atomic representing measures in Theo-rems 10.1 and 10.2. In the subsequent sections we study such representing measures.

10.2 The Moment Cone SmC1 and Its Boundary Points

The following notions play a crucial role in this chapter.

Definition 10.4 The moment cone SmC1 and the moment curve cmC1 are defined by

SmC1 D�s D .s0; s1; : : : ; sm/ W sj D

Z b

at j d�.t/; j D 0; : : : ;m; � 2 MC.Œa; b�/

�;

cmC1 D fs.t/ WD .1; t; t2; : : : ; tm/ W t 2 Œa; b� g

That is, SmC1 is the set of moment sequences s D .s0; s1; : : : ; sm/ of all Radonmeasures on Œa; b�. The curve cmC1 is contained in SmC1, since s.t/ is the momentsequence of the delta measure ıt.

By a slight abuse of notation we consider SmC1 and cmC1 as subsets of RmC1 byidentifying the row vectors s and s.t/ with the corresponding column vector sT ands.t/T in RmC1:

We denote by @SmC1 the set of boundary points, by IntSmC1 the set ofinterior points of SmC1, and by CmC1 the conic hull of cmC1. Since the polynomials1; t; : : : ; tm are linearly independent, the points s.t/ 2 CmC1 span RmC1. Hence theinterior of CmC1 is not empty by Proposition A.33(i).

From Theorem 1.26, applied to E D RŒx�m and X D Œa; b�, it follows that eachs 2 SmC1; s ¤ 0, has a k-atomic representing measure

� DkX

jD1mjıtj ; (10.6)

where k � mC1 and tj 2 Œa; b� for all j. Since� is k-atomic, the points tj are pairwisedistinct and mj > 0 for all j. The numbers tj are called roots or atoms of � and thenumbers mj are the weights of �. We can assume without loss of generality that

a � t1 < t2 < � � � < tk � b: (10.7)


That the measure � in (10.6) is a representing measure of s means that

s DkX

jD1mjs.tj/:

Thus s belongs to the conic hull of the moment curve cmC1. Hence SmC1 D CmC1.By Theorem 1.26(ii), the set of functionals Ls, where s 2 SmC1, is closed in the dualspace E�. This implies that CmC1 is closed in RmC1: We state these results.

Proposition 10.5 The moment cone SmC1 is a closed convex cone in RmC1 withnonempty interior. It is the conic hull CmC1 of the moment curve cmC1.

By Definition A.38, a convex subset B of a cone C is a base of C if for eachu 2 C, u ¤ 0, there exists a unique � > 0 such that �u 2 B. It is easily seen that

Sm D f.1; s1; : : : ; sm/ 2 SmC1g

is a base of the cone SmC1: It is just the set of moment sequences of all probabilitymeasures on Œa; b�.

Clearly, jsjj D jR ba t jd�j � b � a for s 2 Sm. Hence Sm is bounded in RmC1.

Obviously, Sm is a closed subset of the closed cone SmC1. Therefore, the set Sm is aconvex compact base of the moment cone SmC1.

Definition 10.6 Let s 2 SmC1; s ¤ 0. The index ind.�/ of the k-atomicrepresenting measure (10.6) for s is the sum

ind.�/ WDkX

jD1�.tj/; where �.t/ WD 2 for t 2 .a; b/ and �.a/ D �.b/ WD 1:

The index ind.s/ of s is the minimal index of all representing measures (10.6) for s.

The reason why boundary points and interior points are counted differently lies inthe following fact, which enters into many proofs given in this chapter: If t0 2 Œa; b�is a zero with multiplicity k of a polynomial p 2 Pos.Œa; b�/m, then we have k � 2if t0 2 .a; b/, while k D 1 is possible if t0 D a or t0 D b.

Let us recall some further notions. For s 2 SmC1 let Ms denote the set of allrepresenting measures for s, that is, Ms is the set of � 2 MC.Œa; b�/ such that

sj DZ b

at j d�.t/ for j D 0; : : : ;m:

If the set Ms is a singleton, then s is called Œa; b�-determinate.

10.2 The Moment Cone SmC1 and Its Boundary Points 233

The following theorem brings a number of important properties together.

Theorem 10.7 For s 2 SmC1, s ¤ 0, the following statements are equivalent:(i) s is a boundary point of the convex cone SmC1.

(ii) ind.s/ � m.(iii) There exists a p 2 Pos.Œa; b�/m, p ¤ 0, such that Ls. p/ D 0.(iv) Dm.s/ D 0 or Dm.s/ D 0:(v) s is Œa; b�-determinate, that is, Ms is a singleton.

If p is as in (iii), then supp� Z. p/ for the unique measure � 2Ms.

Proof Let fe0; : : : ; emg denote the canonical vector space basis of RmC1.(i)!(iii) Let � 2 Ms. Since s is a boundary point of the cone SmC1, there is a

supporting hyperplane of SmC1 at s (by Proposition A.34(ii)), that is, there exists alinear functional F ¤ 0 on RmC1 such that F.s/ D 0 and F � 0 on SmC1. Thenp.t/ D F.s.t// is a polynomial in t of degree at most m. From s.t/ 2 SmC1 we getp 2 Pos.Œa; b�/. Since f ¤ 0, p is not the zero polynomial. Then

Ls. p/ D L.F.s.t// DZ b

aF.s.t//d�.t/ D

Z b

a.F.e0/C � � � C F.em/t

m/d�.t/

D F.e0/s0 C � � � C F.em/sm D F.s/ D 0:

This proves (iii).(iii)!(i) Let p.t/ D Pm

jD0 ajt j be a polynomial as in (iii). Define a linearfunctional F on RmC1 by F.ej/ D aj, j D 0; : : : ;m. Reversing the precedingreasoning, we get F.s/ D Ls. p/ D 0. Since p is not the zero polynomial, F ¤ 0.Further, F.s.t// D p.t/ � 0 on Œa; b�, since p 2 Pos.Œa; b�/. Therefore, F � 0

on cmC1 and hence on its conic convex hull CmC1 D SmC1: Thus F is a supportingfunctional to SmC1 at s. Hence s is a boundary point of SmC1 by Proposition A.34(ii).

(ii)!(iii) Assume that ind.s/ � m: Let p denote the product of quadratic factors.t� tj/2 for all tj 2 .a; b/ and linear factors .b� t/ and .t � a/ provided that the endpoints b or a are among the points tj, respectively. Clearly, then p 2 Pos.Œa; b�/ anddeg. p/ DPj �.tj/ D ind.s/ � m.

(iii)!(ii) Let p be a polynomial as in (iii). We choose a representing measure(10.6) for which ind .s/ D ind .�/. Since Ls. p/ D 0, Proposition 1.23 yieldsft1; : : : ; tkg D supp� Z. p/, that is, p.tj/ D 0 for j D 1; : : : ; k. Each root tj 2.a; b/ has even multiplicity. (Otherwise p.t/ would change sign at tj. Since p.t/ � 0on Œa; b�, this is impossible.) The only roots of multiplicity 1 are possibly the endpoints a and b. Therefore, counting the roots tj with multiplicities and adding thenumber �.tj/ we obtain ind.s/ DPj �.tj/ � deg. p/ � m:

(iii)$(iv) We carry out the proof in the even case m D 2n. The proof in the oddcase is similar; instead of (10.2) and (10.4) we use (10.3) and (10.5).


Let f .x/ D PniD0 aixi 2 RŒx�n and g.x/ D Pn�1

jD0 bjxj 2 RŒx�n�1. As above we

set Ef D .a0; : : : ; an/T 2 RnC1 and Eg D .b0; : : : ; bn�1/T 2 Rn. The Hankel matricesH2n.s/ and H2n.s/ are positive semidefinite by Theorem 10.1. Therefore, by (10.2),

Ls. f2/ DEf TH2n.s/Ef D kH2n.s/

1=2Ef k2; (10.8)

Ls..b � x/.x � a/g2/ D EqTH2n.s/Eg D kH2n.s/1=2Egk2; (10.9)

where k � k denotes the Euclidean norm of Rd. By (10.4), Pos.Œa; b�/2n consistsof sums of polynomials of the form p D f 2 C .b � x/.x � a/g2. Clearly, for such apolynomial p we have Ls. p/ D 0 if and only if Ls. f 2/ D 0 and Ls..b�x/.x�a/g2/ D0: Therefore, by (10.8) and (10.9), Ls. p/ D 0 for p 2 Pos.Œa; b�/2n implies p D 0

if and only if both matrices H2n.s/ and H2n.s/ are regular, that is, D2n.s/ ¤ 0 andD2n.s/ ¤ 0. Hence (iii) holds if and only if D2n.s/ D 0 or D2n.s/ D 0.

(iii)!(v) Let � 2 Ms. By Proposition 1.23, supp � Z. p/. In particular,this proves the last assertion. Let Z. p/ D fx1; : : : ; xrg. Then � D Pr

iD1 niıxi forsome numbers ni � 0. Since r � deg. p/ � m, there exist Lagrange interpolationpolynomials pj 2 RŒx�m such that pj.xi/ D ıij, i; j D 1; : : : ; r. Then

Ls. pj/ DZ b

apj d� D

rXiD1

nipj.xi/ D nj; j D 1; : : : ; r:

This shows that � is uniquely determined by s and p.(v)!(i) If s is not in @SnC1, then s 2 IntSmC1; hence s has infinitely many

representing measures by Proposition 10.9 below. utWe briefly discuss the preceding theorem. Let s be a boundary point of SmC1.

Then s has a unique representing measure � 2 MC.Œa; b�/: This measure � has theform (10.6) and ind .�/ D ind.s/ � m. If all tj are in the open interval .a; b/, thenk � m

2. If precisely one tj is an end point, then k � mC1

2and if both end points are

among the tj, then k � m2C 1. The case k D m

2C 1 can only happen if m is even and

both a and b are among the tj. Thus, all boundary points of SmC1 can be representedby k-atomic measures with k � m

2C 1:

10.3 Interior Points of SmC1 and Interlacing Propertiesof Roots

The following result is often used in the sequel. Condition (ii) and (iii) therein shouldbe compared with conditions (ii) and (iii) in Theorems 10.1 and 10.2.

10.3 Interior Points of SmC1 and Interlacing Properties of Roots 235

Theorem 10.8 For a sequence s 2 SmC1 the following are equivalent:

(i) s is an interior point of SmC1.(ii) The Hankel matrices Hm.s/ and Hm.s/ are positive definite.

(iii) m D 2n WLs. p2/ > 0 and Ls..b� x/.x� a/q2/ > 0 for p 2 RŒx�n, q 2 RŒx�n�1; p; q¤0:m D 2nC 1 WLs..x � a/p2/ > 0 and Ls..b � x/p2/ > 0 for all p 2 RŒx�n; p ¤ 0.

(iv) Dm.s/ > 0 and Dm.s/ > 0:(v) D0.s/ > 0;D0.s/ > 0;D1.s/ > 0;D1.s/ > 0; : : : ;Dm.s/ > 0;Dm.s/ > 0.

Proof (i)$(iv) is only a restatement of Theorem 10.7 (i)$(iv). The equivalence(ii)$(iv) is a well-known fact from linear algebra. The equivalence of (iii) and(iv) follows from formulas (10.2) and (10.3). (v)!(iv) is trivial. Obviously, s 2IntSmC1 implies that s. j/ WD .s0; : : : ; sj/ 2 IntSjC1 for j D 0; : : : ;m. Therefore,(i)!(v) follows by applying the implication (i)!(ii) to the sequences s. j/. utProposition 10.9 Let s 2 IntSmC1. For each � 2 Œa; b� there exists a representingmeasure �� of s with � as an atom which has index mC 1 or mC 2 if � 2 .a; b/ andindex mC 1 if � D a or � D b.

Proof Clearly, c.�/ is a boundary point, because ind.c.�// � 1. The line throughthe two different points s and c.�/ of SmC1 intersects the boundary of SmC1 in asecond point s0. Then s D �s0 C .1 � �/c.�/ for some � 2 .a; b/. The uniquerepresenting measure �0 of s0 satisfies ind.�0/ � m by Theorem 10.7. It is clearthat � D ��0 C .1� �/ı� is a representing measure of s. Its index is at most mC 2if � 2 .a; b/ and at most m C 1 if � D a or � D b. We have ind.s/ > m, sinceotherwise s would be a boundary point by Theorem 10.7. utDefinition 10.10 A representing measure � of s 2 IntSmC1 is called canonicalif � is of the form (10.6) and ind .�/ � mC 2.

Each representing measure �� from Proposition 10.9 is canonical.The following two propositions are useful for deriving interlacing properties of

roots for different canonical measures of the same moment sequence s 2 IntSmC1:

Proposition 10.11 Suppose that � is a canonical representing measure of s 2IntSmC1 with roots t1 < t2 < � � � < tk and weights m1; : : : ;mk. Let Q� be arepresenting measure of s such that Q� ¤ �. Then we have:

(i) supp Q�\ .tj; tjC1/ ¤ ; for tj; tjC1 2 .a; b/.(ii) If ind .�/ D mC 1, then supp Q� \ .tj; tjC1/ ¤ ; for all tj; tjC1 2 Œa; b�.

(iii) If t1 D a, then supp Q�\ Œt1; t2/ ¤ ;:(iv) If tk D b, then supp Q� \ .tk�1; tk� ¤ ;:Proof

(i) The proof is a modification of the proof of Theorem 10.7 (iii)!(iv).Fix tj; tjC1 2 .a; b/. Let p.t/ denote the product of .t�tjC1/.t�tj/; all factors

.t�ti/2 with ti 2 .a; b/; ti ¤ tj; tjC1, and possibly .t�a/ resp. .b�t/ if a resp. b


are among the roots. Then p.tj/ D 0 for all j. Since ind.�/ � mC 2, we havedeg. p/ � m: (Note that each of the two interior roots tj; tjC1 is counted with 2in the definition of ind .s/, but it appears only with degree 1 in p.) Hence thefunctional Ls applies to p and we obtain

Ls. p/ DZ b

ap d� D

kXjD1

mjp.tj/ D 0:

Assume to the contrary that supp Q� \ .tj; tjC1/ D ;: By the definition of pwe have p.t/ � 0 on Mj WD Œ0; tj� [ ŒtjC1; 1�. Therefore, since 0 D Ls. p/ DRMjp d Q�, Proposition 1.23 implies that supp Q� Z. p/ D ft1; : : : ; tkg. Thus,

Q� D PkiD1 niıti for some numbers ni � 0: Since ind .�/ � m C 2 by

assumption, we have k � m2C 2 and hence k � mC 1, because k is an integer

and m � 1. Hence there exist interpolation polynomials pj of degree deg. pj/ �m satisfying pj.xi/ D ıij, i; j D 1; : : : ; k. Finally, Ls. pj/ D

Rpj d Q� D nj and

Ls. pj/ DRpj d� D mj imply that nj D mj for all j. Thus � D Q�; which is the

desired contradiction.(ii) In the case ind .�/ D m C 1 the preceding reasoning also works for the

end points among the roots. We illustrate this for t1 D a. Let p.t/ denote theproduct of .t � a/.t � t2/, all factors .t � ti/2 with ti 2 .a; b/; ti ¤ tj; tjC1, and.b � t/ if b is a root. Then deg. p/ D m, because ind .�/ D m C 1. Sincep.tj/ D 0 for all j and p.t/ � 0 on Œt2; b�, we can proceed as above and derivethat supp Q� \ .t1; t2/ ¤ ;:

(iii) Assume to the contrary that supp Q� \ Œa; t2/ D ;:First we note that t2 ¤ b. Indeed, otherwise Q� has support fbg and hence

ind. Q�/ D 1 which contradicts the assumption s 2 IntSmC1.Let q be the product of factors .t�t2/, .t�ti/2 if i � 3 and ti ¤ b, and .tk�t/

if tk D b. Since t1 D a, t2 < b, and ind .�/ � mC2, we have deg.q/ � m. Thepolynomial q vanishes exactly at t2; : : : ; tk. Hence Ls.q/ D

R ba q d� D m1q.a/.

By construction, q � 0 on Œt2; b� and therefore Ls.q/ DR ba q d Q� D R b

t2q d Q� �

0. But q.a/ < 0; so we obtain a contradiction.(iv) follows in a similar manner as (iii). utProposition 10.12 Let � and �0 be two different canonical representing measuresof s 2 IntSmC1 with roots t1 < t2 < � � � < tk and t01 < t02 < � � � < t0l and weightsmi and m0

j, respectively.

(i) The roots ti and t0j contained in .a; b/ strictly interlace.(ii) Suppose that t1 D t01 D a. Then m1 ¤ m0

1. Further, m01 > m1 if and only if

t02 > t2:(iii) If tk D t0l D b, then mk ¤ m0

l and we have m0l > mk if and only if tl�1 > t0k�1:

10.4 Principal Measures of Interior Points of SmC1 237

Proof

(i) By assumption � and �0 have index mC 1 or mC 2. Considering the even andthe odd case separately we see that the numbers of their roots in .a; b/ differby at most one. By Proposition 10.11 (i) and (iii), the smallest inner roots of� and �0 are different. Therefore it follows from Proposition 10.11(i) that theinner roots of � and �0 strictly interlace.

(ii) The crucial step of the proof is to show that t02 > t2 implies m01 > m1: Let q be

the polynomial defined in the proof of Proposition 10.11(iii). Let us recall thatdeg.q/ � m, q � 0 on Œt2; b� and q vanishes exactly at the roots t2; : : : ; tk. Theroot t02 cannot be equal to all roots ti; i � 3. (This would imply that t02 D t3 D band k D 3, so ind.�0/ D 2 and ind.�/ D 4, which is impossible.) Therefore,R bt2q d�0 > 0 and

m1q.a/ DZ b

aq d� D Ls.q/ D

Z b

aq d�0 D m0

1q.a/CZ b

t2

q d�0 > m01q.a/:

Since q.a/ < 0, the latter yields m01 > m1.

By interchanging the role of � and �0 it follows that t2 > t02 implies m1 >m01:

Finally, we show that m1 ¤ m01: Assume to the contrary that m1 D m0

1: Thenwe have t2 D t02 by the preceding. Since inner roots strictly interlace by (i), thiscan only happen if t2 D t02 D b: But since � and �0 have the same total masss0 and m1 D m0

1; it follows that � D �0. This contradicts our assumption.(iii) is proved in a similar manner as (ii). utCorollary 10.13 Let s 2 IntSnC1. For each � 2 .a; b/ there exists a uniquecanonical representing measure �� of s with atom �.

Proof The existence has been already stated in Proposition 10.9, so it remainsto verify the uniqueness. If there were two different such measures, their rootscontained in .a; b/ would strictly interlace by Proposition 10.12(i). Since bothmeasures have the same root � 2 .a; b/, this is impossible. ut

10.4 Principal Measures of Interior Points of SmC1

Now we turn to the minimal possible index m C 1 for interior points of SmC1.Throughout this section we suppose that s 2 IntSmC1.

Definition 10.14 A representing measure � of the form (10.6) for s is called

� principal if ind .�/ D mC 1,� upper principal if it is principal and b is an atom of �,� lower principal if it is principal and b is not an atom of �.


That is, for principal measures the index is equal to the number of prescribedmoments. The representing measures constructed in the proof of Proposition 10.9for � D 0 and for � D 1 are principal. We will return to these measures later.

Given a fixed sequence s D .s0; : : : ; sm/ 2 SmC1 we define

sCmC1 D sup

�2Ms

Z b

axmC1 d�.x/ D sup

.s0;:::;sm;smC1/2SmC1

smC1 (10.10)

s�mC1 D inf

�2Ms

Z b

axmC1 d�.x/ inf

.s0;:::;sm;smC1/2SmC1

smC1: (10.11)

Since Ms is weakly compact (by Theorem 1.19) andR 10xmC1 d� is a continuous

function on the compact space Ms, the supremum in (10.10) and the infimumin (10.11) are attained. Thus, sC

mC1 is the maximum and s�mC1 is the minimum of

the moment smC1 over the set Ms of all measures which have the given momentss0; : : : ; sm: These extremal values sC

mC1 and s�mC1 play a crucial role in our approach

to the principal measures �˙, but they are also of interest in themselves.Let s˙ denote the sequence .s0; : : : ; sm; smC1/ 2 SmC2. Since sC

mC1 is a maximumand s�

mC1 is a minimum, sC and s� are not in IntSmC2. Hence s˙ belongs to theboundary of SmC2, so by Theorem 10.7 it has a unique representing measure �˙ ofindex ind .�˙/ � m C 1: Obviously, �C and �� are also representing measurefor the sequence s. Since s 2 IntSmC1, ind .s/ > m by Theorem 10.7. Henceind .�˙/ D mC 1, that is, �C and �� are principal representing measures for s:

The next proposition characterize the numbers smC1 in terms of determinants ofHankel matrices.

Proposition 10.15 Let Qs D .s0; : : : ; sm; smC1/ 2 SmC2: Then sCmC1 and s�

mC1 arethe unique numbers smC1 satisfying DmC1.Qs/ D 0 and DmC1.Qs/ D 0, respectively.Further, we have sC

mC1 > s�mC1 and

DmC1.sC/ D 0; DmC1.sC/ > 0; DmC1.s�/ D 0; DmC1.s�/ > 0: (10.12)

Proof We develop DmC1.Qs/ and DmC1.Qs/ by the last row and obtain

DmC1.Qs/ D smC1Dm�1.s/C cC; DmC1.Qs/ D �smC1Dm�1.s/C c� (10.13)

for some numbers c˙ depending only on the given moments s0; : : : ; sm.We prove that s�

mC1 is the unique number smC1 for which DmC1.Qs/ D 0. Since s 2IntSmC1, we have Dm�1.s/ > 0 by Theorem 10.7 (i)$(iv). Therefore, by (10.13),DmC1.Qs/ is a strictly increasing function of smC1. Let us take some Qs 2 IntSmC2.Then DmC1.Qs/ > 0 and DmC1.Qs/ > 0 by Theorem 10.8. Now we decrease smC1untill s�

mC1 to obtain DmC1.Qs/ D 0. Then DmC1.Qs/ increases by (10.13) and remainspositive. Hence s�

mC1 is the lower bound of the numbers smC1 for which DmC1.Qs/ �0. Since Hm.s/ is positive definite because s 2 IntSmC1, s�

mC1 is also the lower

10.4 Principal Measures of Interior Points of SmC1 239

bound of the numbers smC1 such that HmC1.Qs/ � 0, or equivalently, Qs 2 SmC2.Thus, s�

mC1 is the number defined by (10.11).Similarly it follows that the number sC

mC1 from (10.10) is uniquely determinedby the equation DmC1.Qs/ D 0: Moreover, the proof shows that sC

mC1 > s�mC1: ut

Before we continue we note an interesting by-product of the preceding.

Corollary 10.16 Let s D .s0; : : : ; sm/ 2 IntSmC1 and let smC1 be a real number.Then Qs D .s0; : : : ; sm; smC1/ 2 SmC2 if and only if sC

mC1 � smC1 � s�mC1.

Proof The only if part is clear from the definition of smC1. Conversely, suppose thatsCmC1 � smC1 � s�

mC1. Then smC1 D �sCmC1 C .1 � �/s�

mC1 for some � 2 Œ0; 1�.Therefore, ��CC .1� �/�� is a representing measure for Qs, so that Qs 2 SmC2. ut

Next we show that the two measures �˙ defined above are the only prinicipalrepresenting measures of s. Let � be an arbitrary principal measure for s. Clearly,Qs WD .s0; : : : ; sm; smC1/ 2 SmC2, where smC1 WD

R ba xmC1 d�. Since ind .�/ D

m C 1, DmC1.Qs/ D 0 or DmC1.Qs/ D 0 by Theorem 10.7 (ii)!(iv). Hence, byProposition 10.15, smC1 D s�

mC1 or smC1 D sCmC2, which implies that Qs D s� or

Qs D s�: Since �˙ is the unique representing measure of s˙, we conclude that� D �� or � D �C.

We denote by tj the roots and bymj the weights of�˙ and assume that the roots

are ordered as in (10.7). Since ind .�˙/ D mC1, it follows from Proposition 10.11that the roots of �C and�� are strictly interlacing. To describe their location furtherwe distinguish between the even and odd cases.

Case m D 2n:Since both measures �˙ have index 2n C 1, they have exactly n roots containedin .a; b/ and one end point a or b as root. Let f .x/ D Pn

iD0 aixi 2 RŒx�n. SettingEf D .a0; : : : ; an/T 2 RnC1 and using (10.3) we compute

Ef T HmC1.sC/Ef D LsC..b � x/f 2/ DZ b

a.b � x/f .x/2 d�C D

nC1XjD1

mCj .b � tCj /f .t

Cj /

2:

Since DmC1.sC/ D 0, the Hankel matrix HmC1.sC/ has a nontrivial kernel. Hencethere exists an f ¤ 0 such that LsC..b� x/f 2/ D 0, that is, .b� tCj /f .tCj / D 0 for allj D 1; : : : ; nC 1. Since there are nC 1 roots and deg. f / � n, this is only possibleif tCnC1 D b. A similar reasoning using HmC1.s�/ instead of HmC1.sC/ shows thatt�1 D a. Thus, �C is upper principal and �� is lower principal.

Since s has only two principal measures, we conclude that �C and �� are theprincipal representing measures �� from Proposition 10.9 for � D b and � D a; thatis, �C D �b and �� D �a:

Case m D 2nC 1:Since ind .�˙/ D 2n C 2, �˙ has either n C 1 inner roots or n inner roots andboth end points as roots. Thus �C has k � nC 1 roots. Let f 2 RŒx�n: Using (10.2)


we derive

Ef T HmC1.sC/Ef D LsC..b � x/.x � a/f 2/ DkX

jD1mC

j .b � tCj /.tCj � a/f .tCj /

2:

As in the even case, HmC1.sC/ has a nontrivial kernel, so there is a polynomialf ¤ 0 such that LsC..b � x/.x � a/f 2/ D 0. Then .b � tCj /.t

Cj � a/f .tCj / D 0 for

j D 1; : : : ; k. Since k � nC1 and deg. f / � n, one root of �C must be an end point.Therefore, since ind .�C/ D 2nC 2, it follows that tC1 D a, tCnC2 D b, k D nC 2.

By Proposition 10.11, the roots of �C and �� strictly interlace. Therefore, tCj <

t�j and t�i < tCiC1. Hence �C has both end points as roots, while �� has only innerroots. As in the even case, �C is upper principal and �� is lower principal.

In contrast to the even case, �C is equal to both principal measures �� fromProposition 10.9 for � D b; a; that is, �C D �b D �a: The second principalmeasure �� has only inner roots and it is not obtained from Proposition 10.9.

We illustrate the location of roots tj of the principal measures �˙ in the evenand odd cases by the following scheme:

m D 2n; �C W a < tC1 < tC2 < � � � < tCnC1 D b; (10.14)

m D 2n; �� W a D t�1 < t�2 < � � � < t�nC1 < b; (10.15)

m D 2nC 1; �C W a D tC1 < tC2 < � � � < tCnC1 < tCnC2 D b; (10.16)

m D 2nC 1; �� W a < t�1 < t�2 < � � � < t�nC1 < b: (10.17)

Further, from Proposition 10.12 we obtain

m D 2n W a D t�1 < tC1 < t�2 < tC2 < � � � < tCn < t�nC1 < tCnC1 D b;

(10.18)

m D 2nC 1 W a D tC1 < t�1 < tC2 < t�2 < � � � < tCnC1 < t�nC1 < tCnC2 D b:

(10.19)

The following theorem summarizes some of the preceding results.

Theorem 10.17 Let s be an interior point of SmC1. Then �C is the unique upperprincipal representing measure and �� is the unique lower princical representingmeasure for s. The roots of �C and �� are stricly interlacing.

Now we define a distinguished canonical representing measure �� and a relatedpolynomial q� 2 Pos.Œa; b�/ for each � 2 Œa; b�. Both will play a crucial role in thenext sections.

Definition 10.18 Let s 2 IntSmC1. For � 2 .a; b/, �� is the unique canonicalmeasure of s which has � as a root (see Corollary 10.13). For � D a or � D b, �� is

10.5 Maximal Masses and Canonical Measures 241

the unique principal measure of s which has � as a root, that is, �a D ��; �b D �Cfor m D 2n and �a D �b D �C for m D 2nC 1.

If � 2 .a; b/ is root of a principal measure �˙, then �� D �˙ has index mC 1.Note that �a resp. �b is the only principal measure with root a resp. b, but there aremany canonical measures with root a resp. b, see Theorems 10.25 and 10.26 below.

Definition 10.19 For � 2 Œa; b� let the polynomial q�.x/ be the product of factors.x � tj/2 for all roots tj 2 .a; b/, tj ¤ �, of the measure �� and .x � a/ resp. .b � x/if a resp. b is a root of �� .

Lemma 10.20 Let � 2 Œa; b� and let �; t1; : : : ; tk denote the roots of �� . Then, up toa constant positive factor, q� is the unique polynomial q 2 Pos.Œa; b�/ such that

q.�/ > 0; deg.q/C �.�/ � ind.��/ and q.tj/ D 0 for j D 1; : : : ; k: (10.20)

Proof From its definition it is clear that q� has these properties. Since in Defini-tion 10.19 no factor was taken for the root �, we even have deg.q�/ C �.�/ Dind.��/.

Let Qq be another such polynomial. Since Qq 2 Pos.Œa; b�/; its zeros in .a; b/ haveeven multiplicities. Hence we conclude that deg.q�/ � deg.Qq/: But

deg.Qq/C �.�/ � ind.��/ D deg.q�/C �.�/

by (10.20) implies that deg.Qq/ � deg.q�/. Thus deg.Qq/ D deg.q�/. Since Qq.�/ > 0

and q�.�/ > 0, it follows that Qq is a positive multiple of q� . ut

10.5 Maximal Masses and Canonical Measures

Let s 2 SmC1 and � 2 Œa; b�. Then we define

s.�/ D sup f�.f�g/ W � 2Msg; (10.21)

that is, s.�/ is the supremum of masses at � of all representing measures of s. Wewill say that a measure � 2Ms has maximal mass at � if �.f�g/ D s.�/.

Further, we need the following number

s.�/ WD inf fLs. p/ W p 2 Pos.Œa; b�/m; p.�/ D 1 g (10.22)

D inf

�Ls.q/

q.�/W q 2 Pos.Œa; b�/m

�; (10.23)

where c0WD C1 for c � 0. The equality in (10.23) is easily verified.


For arbitrary � 2Ms and p 2 Pos.Œa; b�/m, p.�/ D 1; we obviously have

Ls. p/ DZ b

ap d� � �.f�g/p.�/ D �.f�g/:

Taking the infimum over p and the supremum over �, we derive

s.�/ � s.�/: (10.24)

Theorem 10.21 Suppose that s 2 IntSmC1. Let � 2 Œa; b�.Then the supremum in (10.22) and the infimum in (10.23) are attained and

s.�/ D s.�/ > 0:

The measure �� from Definition 10.19 is the unique representing measure of swhich has maximal mass s.�/ at �.

The infimum in (10.23) is attained at the polynomial q� from Definition 10.19. If� is not an inner root of a principal measure of s, then q� is up to a positive multiplethe only polynomial p 2 Pos.Œa; b�/m for which the infimum in (10.23) is attained.

Proof We denote the roots and masses of �� by t0 D �; t1; : : : ; tk and m0; : : : ;mk.For the polynomial q� from Definition 10.19 we have deg.q�/ � ind .��/��.�/ Dm by (10.20), so Ls applies to q� and we get

Ls.q�/ DZ b

aq� d�� D

kXjD0

mjq�.tj/ D m0q�.�/:

Therefore, since q� 2 Pos.Œa; b�/m, we obtain

s.�/ � m0 D Ls.q�/

q�.�/� inf

�Ls.q/

q.�/W q 2 Pos.Œa; b�/m

�D s.�/:

Combined with (10.24) we conclude that we have equality throughout, that is,

��.f�g/ D m0 D s.�/ D Ls.q�/

q�.�/D s.�/: (10.25)

The first equalities of (10.25) mean that the measure �� has maximal mass at �. Thelast equality of (10.25) says that the infimum in (10.23) is attained at q� .

Let � be an arbitrary representing measure for s which has maximal mass s.�/at �. Then � WD � � s.�/ı� is a positive measure satisfying

Z b

aq�d� D

Z b

aq�d� � s.�/q�.�/ D Ls.q�/� s.�/q�.�/ D 0 (10.26)


by (10.25). Since q� 2 Pos.Œa; b�/m, (10.26) implies that supp� Z.q�/ byProposition 1.23. Hence supp � f�g [ Z.q�/. Therefore, � is atomic and allatoms of � are roots of �� . Thus ind.�/ � ind.��/ and � has � as an atom, since�.f�g/ D s.�/ > 0. If � 2 .a; b/, then �� is canonical, hence is �, and therefore� D �� by Corollary 10.13. If � D a or � D b, then �� , hence �, is principal. Sincethere is only one principal measure with root at a given endpoint, we obtain � D ��in this case as well.

Now let q 2 Pos.Œa; b�/m be another polynomial for which the infimum s.�/ in(10.23) is attained, so that s.�/q.�/ D Ls.q/ and q.�/ > 0. Then, by (10.25),

m0q.�/ D s.�/q.�/ D Ls.q/ DkX

jD0mjq.tj/ D m0q.�/C

kXjD1

mjq.tj/:

Since q 2 Pos.Œa; b�/m and mj > 0, we conclude that q.tj/ D 0 for j D 1; : : : ; k.Suppose that � is not an inner root of a principal measure for s. Then we have

ind.��/ D mC 2 if � 2 .a; b/ and ind.��/ D mC 1 if � D a or � D b. In bothcases, m D ind.��/ � �.�/, so that deg.q/ C �.�/ � ind .��/. Thus q satisfies(10.20). Hence q is a positive constant multiple of q� by Lemma 10.20. utRemark 10.22

1. The preceding proof shows that for a polynomial q 2 Pos.Œa; b�/m the infimum in(10.23) is attained if and only if q.tj/ D 0 for all roots tj ¤ � of �� and q.�/ > 0.Let � be an inner root of a principal measure of s. Then deg.q�/ D m� 1 and theinfimum in (10.23) is attained at q 2 Pos.Œa; b�/m if and only if q D fq� for someconstant or linear polynomial f 2 Pos.Œa; b�/.

2. Let s be a boundary point of SmC1. Then s has a unique representating measure� by Theorem 10.7, so s.�/ is the corresponding weight if � is a root of � ands.�/ D 0 if � is not a root of �. ıWe derive two important consequences of the preceding theorem.

Corollary 10.23 Let � be a canonical representing measure of s 2 IntSmC1 withroots tj; j D 1; : : : ; k: Then � has maximal mass at each root tj contained in .a; b/.If � is principal, � has maximal mass at all roots.

Proof If � is canonical and tj 2 .a; b/ is a root of �, then �tj D � byCorollary 10.13 and Definition 10.18. If � is principal and if tj D a or tj D b,then also �tj D � by Definition 10.18. In both cases � has maximal mass at tj byTheorem 10.21. utCorollary 10.24 For each s 2 SmC1; s ¤ 0; there is a representing measure � of ssuch that ind .�/ � mC 1 and � has maximal mass at each root.

Proof First let s 2 @SmC1. Then, by Theorem 10.7, ind.�/ � m and s is Œa; b�-determinate, so � obviously has maximal mass at all its roots. If s 2 IntSmC1,each principal measure has the desired properties by Corollary 10.23. ut


The following two theorems deal with canonical measures having prescribedmasses at the end points. Because of the different behaviour of principal measuresat end points we distinguish between the even and odd cases.

Theorem 10.25 (Odd case m D 2nC 1) Suppose that � and � 0 are numbers suchthat 0 < � < s.a/ and 0 < � 0 < s.b/. Then there exist unique canonical measures�a.�/ and �b.� 0/ for s which have masses � and � 0 at a and b, respectively.

Let a D ta1.�/ < ta2.�/ < � � � < tak.�/ and tb1.�0/ < tb2.�

0/ < � � � < tbl .�0/D b

denote the roots of �.�/ and �.� 0/, respectively. Each root tj.�/ is a strictlyincreasing continuous function of � on the interval .0; s.a//, while tj.� 0/ is a strictlydecreasing continuous function of � 0 on .0; s.b//. Further, k D l D nC 2 and

t�j�1 D lim�!C0 t

aj .�/; tCj D lim

�!s.a/�0taj .�/; j D 2; : : : ; nC 2;

a D tC1 D ta1.�/ <t�1 < ta2.�/ < tC2 < t�2 < � � � < t�nC1 < tanC2.�/ < tCnC2 D b;

tCj D lim� 0!C0 t

bj .�

0/; t�j D lim� 0!s.b/�0

tbj .�0/; j D 1; : : : ; nC 1;

a D tC1 < tb1.�0/ < t�1 < tC2 < � � � < tCnC1 < tbnC1.� 0/ < t�nC1 < tbnC2.� 0/ D tCnC2 D b:

Proof We carry out the proof for the end point a and define a sequence r.�/ WD.sj � �a j/mjD0. Since �C has the root a by (10.16) and maximal mass at tC1 D a byCorollary 10.23,�C��ıa is a positive measure. Obviously, its moments are sj��aj,so r.�/ is a moment sequence and

Lr.�/. p/ D Ls. p/� �p.a/ D .s.a/� �/p.a/CnC2XjD2

mjp.tCj /; p 2 RŒx�m:

Since s.a/�� > 0, Lr.�/. p/ D 0 implies that p.tCj / D 0 for all j D 1; : : : ; nC2, thatis, Lr.�/. p/ D 0 is equivalent to Ls. p/ D 0. Hence, by Theorem 10.8, s 2 IntSmC1implies that r.�/ 2 IntSmC1. Thus, r.�/ has a lower principal measure�.�/� withroots tj.�/ written as a < t2.�/ < � � � < tnC2.�/ < b: Then �.�/ WD �.�/� C �ıa isa canonical representing measure of s with roots

a D t1.�/ < t2.�/ < � � � < tnC2.�/ < b:

Since t1.�/ D tC1 D a and � < s.a/ D �C.fag/, Proposition 10.12(ii) applies andyields t2.�/ < tC2 : Applying Proposition 10.12(i) by using this fact we obtain theinterlacing inequalities stated in the theorem.

Suppose that Q�.�/ is an arbitrary canonical measure for s with mass � at a. ThenQ�.�/ � �ıa is a lower principal measure for r.�/. Therefore, Q�.�/ � �ıa D �.�/�,which yields Q�.�/ D �.�/:

Suppose that 0 < � 0 < � 00 < s.a/. Then t2.� 0/ < t2.� 00/ by Propo-sition 10.12(ii). By the strict interlacing of roots (proposition 10.12(i)) we get


tj.� 0/ < tj.� 00/ for j D 2; : : : ; n C 1. Thus, tj.�/ is a strictly increasing functionof � on .0; s.a//.

Finally, we prove the continuity of tj.�/ and the limit equalities in the theorem.Since tj.�/ is increasing, all one-sided limits lim�!�0˙ tj.�/ for �0 2 .0; s.a//,lim�!C0 tj.�/, and lim�!s.a/�0 tj.�/ exist. It only remains to show the correspond-ing equalities. As a sample we assume the contrary for some �0 2 .0; s.a// and jand choose ˛; ˇ such that

tCj�1 < t�j�1 � lim�!�0�0

tj.�/ < ˛ < ˇ < lim�!�0C0

tj.�/ � tCj : (10.27)

Let � 2 Œ˛; ˇ�. By Corollary 10.13, there is a canonical measure �� for s whichhas � as a root. Then �� WD ��.fag/ is in Œ0; s.a/�. If �� < s.a/, then �� D �Cby Theorem 10.21, which contradicts (10.27). We prove that �� > 0. Assume tothe contrary that �� D 0. Since ind.��/ D m C 2 D 2n C 3, then b must bea root of �� . Let �1 < �2 < : : : ; < �nC1 < �nC2 D b be the roots of �� . Since�nC2 D tCnC2 D b and ��.fbg/ < �C.fbg/ again by Theorem 10.21, it followsfrom Proposition 10.12(iii), applied with � D �� and �0 D �C, that tCnC1 < �nC1.Proposition 10.11(iv), applied with � D �� and Q� D ��; yields �nC1 < t�nC1. ThustCnC1 < �nC1 < t�nC1. Therefore, by the interlacing property in Proposition 10.12(i)we get t�j�1 < � < tCj < �j < t�j . Thus �� has no root between the inner roots � and�j of �� . This contradicts Proposition 10.11(i) and proves that �� > 0. Thus we haveshown that �� 2 .0; s.a//.

By construction, � D tj.��/. From the strict monotonocity of the function tj.�/and (10.27) it follows that �� D �0. Then � D tj.�0/. Since � 2 Œ˛; ˇ� was arbitrary,this is impossible. utTheorem 10.26 (Even Case m D 2n) Let � be a number such that 0 < � < s.a/.Then there exists a unique canonical measure �.�/ for s which has mass � at a. Leta D t1.�/ < t2.�/ < � � � < tk.�/ denote the roots of �.�/. Then k D n C 2. Eachroot tj.�/ is a strictly increasing continuous function of � on the interval .0; s.a//and

tCj�1 D lim�!C0 tj.�/; t�j D lim

�!s.a/�0tj.�/; j D 2; : : : ; nC 1;

a D t�1 Dt1.�/ < tC1 < t2.�/ < t�2 < tC2 < : : : < tnC1.�/ < t�nC1 < tnC2.�/DtCnC1 D b:

The proof of Theorem 10.26 follows a similar pattern as the proof of Theo-rem 10.25; we omit the details.

However, there is an essential difference between Theorems 10.25 and 10.26: Inthe odd case Theorems 10.25 gives parametrizations of all canonical measures ��when � is not a root of a principal measure in terms of the masses � 2 .0; s.a// at aand � 0 2 .0; s.b// at b. In the even case this is not true, since the measures �� with� contained in some interval .t�j ; t

Cj / are not obtained in Theorem 10.26.


10.6 A Parametrization of Canonical Measures

Let us set J WD [jJj and K WD [lKl, where Jj and Kl are the intervals defined by

m D 2n W Kj D .tCj ; t�jC1/ for jD1; : : : ; n; Jl D .t�l ; tCl / for lD1; : : : ; nC 1;m D 2nC 1 W Kj D .tCj ; t�j / and Jj D .t�j ; tCjC1/ for j D 1; : : : ; nC 1:

That is, J[K is the set of points � 2 Œa; b�which are not roots of a principal measure,or equivalently, for which the measure �� from Definition 10.18 has index m C 2.Our aim in this section is to give a parametrization of measures �� , � 2 J [ K, interms their smallest inner root.

Let us denote the roots of �� by tj.�/ and assume that they are numbered inincreasing order. Using the interlacing results of Propositions 10.11 and 10.12 andarguing similarly as in the proof of Theorem 10.25 the following properties arederived.

Even Case: m D 2n:Let � 2 J1 D .a; tC1 /. The roots of �� are t1.�/ D � and t j.�/ 2 Jj, j D 1; : : : ; nC 1.Each root t j.�/ is a strictly increasing continuous function on J1 and we have

� 2 J1 W lim�!t�1 C0 t

j.�/ D t�j ; lim�!tC1 �0

t j.�/ D tCj ; j D 1; : : : ; nC 1; (10.28)

a D t�1 < t1.�/ < tC1 < t�2 < t2.�/ < tC2 < � � � < t�nC1 < tnC1.�/ < tCnC1 D b:

Let � 2 K1 D .tC1 ; t�2 /. Then �� has the roots t1.�/ D a, t2.�/ D �, t jC1.�/ 2 Kj

for j D 1; : : : ; n, and tnC2.�/ D b: Each function tj.�/ is strictly increasing andcontinuous on K1. Further,

� 2 K1 W lim�!tC1 C0

t j.�/ D tCj�1 ; lim�!t�2 �0 t

j.�/ D t�j ; j D 2; : : : ; nC 1; (10.29)

a D t�1 Dt1.�/ < tC1 < t2.�/ < t�2 < : : : < tCn < tnC1.�/ < t�nC1 < tnC2.�/ D tCnC1 D b:

Odd Case: m D 2nC 1:Let � 2 K1 D .a; t�1 / D .tC1 ; t�1 /. The measure �� has roots t1.�/ D �, t j.�/ 2 Kj

for l D 1; : : : ; n C 1 and tnC2.�/ D b. Each root tl.�/ is a strictly increasing andcontinuous function on K1 and

� 2 K1 W lim�!tC1 C0

t j.�/ D tCj ; lim�!t�1 �0 t

j.�/ D t�j ; j D 1; : : : ; nC 1; (10.30)

a D tC1 < t1.�/ < t�1 < tC2 < � � � < tCnC1 < tnC1.�/ < t�nC1 < tnC2.�/ D tCnC2 D b:

10.6 A Parametrization of Canonical Measures 247

Suppose that � 2 J1 D .t�1 ; tC2 /. Then �� has roots t1.�/ D a, t2.�/ D �, and

t jC1.�/ 2 Jj for j D 1; : : : ; n C 1: Each t j.�/ is a strictly increasing continuousfunction on J1 and

� 2 J1 W lim�!t�1 C0 t

j.�/ D t�j ; lim�!tC2 �0

t j.�/ D tCjC1; j D 1; : : : ; nC 1; (10.31)

a D tC1 D t1.�/ < t�1 < t2.�/ < tC2 < t�2 < � � � < t�nC1 < tnC2.�/ < tCnC2 D b:

The preceding gives a continuous parametrization of the roots of the canonicalmeasures �� for � 2 J [ K in terms of their first inner root � contained in J1resp. K1. Further, in all these cases the limits (10.28)–(10.31) show the one-sidedcontinuity of the functions t j.�/ at the corresponding inner roots of the principalmeasure �˙.

Fix an interval Ji and let � 2 Ji. Recall that the polynomial q� from Defini-tion 10.19 is a product of linear and quadratic factors involving the roots t j.�/ of �� .This definition implies that the above parametrization of roots yields a continuousparametrization of the polynomials q� on Ji. Here the vector space RŒx�m is equippedwith some norm. Let Ji D .�C; ��/. Then �C and �� are roots of principalmeasures. From the limits (10.28)–(10.31) we conclude that each one-sided limitp�

˙D lim�!�

˙˙0 q� exists and gives a polynomial p�

˙2 Pos.Œa; b�/m which

is positive at �˙ and vanishes at the other roots of the corresponding principalmeasure. Therefore, by the remarks after the proof of Theorem 10.21, the infimumin (10.23) is attained for p� at � and for p�

˙at �˙, that is, we have s.t/ D Ls. pt/

pt.t/

for all t 2 Œ�C; ��. Thus we have a continuous parametrization of minimizingpolynomials for (10.23) on the closed interval Ji D Œ�C; ��. (Recall that theminimizing polynomial for (10.23) is uniquely determined up to a constant factorfor points in Ji, but not for inner roots of principal measures.) Therefore, since thelinear functional Ls on the finite-dimensional spaceRŒx�m is continuous, the function s.t/ D Ls. pt/

pt.t/is continuous on the closed interval Ji. Verbatim the same proof yields

the continuity of s on the closure Kj of each interval Kj. Further, each inner root isa common end point of some Ji and Kj. Hence s is continuous on Œa; b�. We statethis as

Theorem 10.27 For each s 2 IntSmC1 the function s.t/ D s.t/ is continuouson the interval Œa; b�.

In the preceding proof a continuous parametrization of minimizing polynomials onintervals Ji and Kj was crucial. The following simple example shows that there isno continuous parametrization of minimizing polynomials on Œa; b�.

Example 10.28 Let m D 2. Then a D t�1 < tC1 < t�2 < tC2 D b. Let us takenumbers � 2 J1 D .a; tC1 / and � 2 K1 D .tC1 ; t�2 /. Clearly,

q�.x/ D .x � t2.�//2; q�.x/ D .b � x/.x � a/:


Let p� and p� be minimizing polynomials for (10.23) at � and �, respectively. Thenp� and p� are constant multiples of q� and q� , respectively. Since t2.�/ ! tC2 D bas � ! tC1 � 0, all possible limits of these polynomials are of the form

lim�!tC1 �0

p�.x/ D c1.x � b/2; lim�!tC1 C0

p�.x/ D c2.b� x/.x � a/

with c1 > 0 and c2 > 0. These limits are different minimizing polynomials for s.t

C1 /. This shows that at the inner root tC1 of �C one cannot have two-sided

continuity of minimizing polynomials. ı

10.7 Orthogonal Polynomials and Maximal Masses

Throughout this section, we assume that s 2 IntSmC1.First we define and develop four sequences of orthogonal polynomials. Put

Pk.x/ D

ˇˇˇˇˇˇˇ

s0 s1 s2 : : : sks1 s2 s3 : : : skC1s2 s3 s4 : : : skC2: : :

sk�1 sk skC1 : : : s2k�11 x x2 : : : xk

ˇˇˇˇˇˇˇ

for 2k � 1 � m;

Pk.x/ D

ˇˇˇˇˇˇˇ

r0 r1 r2 : : : rkr1 r2 r3 : : : rkC1r2 r3 r4 : : : rkC2: : :

rk�1 rk rkC1 : : : r2k�11 x x2 : : : xk

ˇˇˇˇˇˇˇ

for 2kC 1 � m;

Qk.x/ D

ˇˇˇˇˇˇˇ

s1 � as0 s2 � as1 : : : skC1 � asks2 � as1 s3 � as2 : : : skC2 � askC1s3 � as2 s4 � as3 : : : skC3 � askC2: : :

sk � ask�1 skC1 � ask : : : s2k � as2k�11 x : : : xk

ˇˇˇˇˇˇˇ

for 2k � m;

Qk.x/ D

ˇˇˇˇˇˇˇ

bs0 � s1 bs1 � s2 : : : bsk � skC1bs1 � s2 bs2 � s3 : : : bskC1 � skC2bs2 � s3 bs3 � s4 : : : bskC2 � skC3: : :

bsk�1 � sk bsk � skC1 : : : bs2k�1 � s2k1 x : : : xk

ˇˇˇˇˇˇˇ

for 2k � m;

10.7 Orthogonal Polynomials and Maximal Masses 249

where we have set

rj WD .aC b/sjC1 � sj � sjC2 for j D 1; : : : ;m � 2:

The coefficients of xk in these determinants are D2k�2;D2k�2;D2k�1; and D2k�1,respectively. Since s 2 IntSmC1, they are positive by Theorem 10.8. Hence eachof the above polynomials has degree k with positive leading coefficient.

Further, we consider the following four sequences

s D .sj/mjD0; .b � E/.E � a/s D .rj/m�2jD0 ; (10.32)

Es � as D .sjC1 � asj/m�1jD0 ; bs� Es D .bsj � sjC1/m�1

jD0 : (10.33)

Their Hankel matrices are given by the formulas at the beginning of Sect. 10.1.Since s 2 IntSmC1, the corresponding Hankel matrices are positive definite byTheorem 10.8, hence are the sequences in (10.32) for even m and in (10.33) forodd m.

The polynomials Pk;Pk;Qk;Qk are orthogonal polynomials for the moment

sequences (10.32)–(10.33). That is, for any polynomial f 2 RŒx�k�1 we have

Ls.Pk f / D L.b�E/.E�a/s.Pk f / D LEs�as.Qkf / D Lbs�Es.Qk f / D 0: (10.34)

A simple verification can be given in a similar manner as in Sect. 5.1.Suppose that � is a representing measure for s. Then the moment sequences

.b�E/.E� a/s, Es� as, and bs�Es are represented by .b� t/.t� a/d�, .t� a/d�,and .b� t/d�, respectively. Hence the above families of polynomials are orthogonalwith respect to the corresponding measures. That is, for all k ¤ j we have

Z b

aPk.t/Pj.t/ d� D

Z b

aPk.t/Pj.t/ .b � t/.t � a/d� D 0 (10.35)

Z b

aQ

k.t/Q

j.t/ .t � a/d� D

Z b

aQk.t/Qj.t/ .b � t/d� D 0:

It should be emphasized that the polynomial Pk is defined if 2k� 1 � m, while Pj isonly if 2jC 1 � m. That is, Pj is not defined for the largest index of Pk.

Now we turn to the orthonormal polynomials. Let m D 2n or m D 2nC 1. Fork D 0; : : : ; n and l D 0; : : : ; n � 1 we define

pk.x/ D Pkp

D2k�2D2k; pl.x/ D

PlqD2lD2lC2

; (10.36)

qk.x/ D Q

kpD2k�1D2kC1

; qk.x/ DQkq

D2k�1D2kC1; (10.37)


where the determinantsDi and Di with negative i are set to 1. All determinantsDj;Dj

occuring in (10.36)–(10.37) are positive by Theorem 10.8, since s 2 IntSmC1.Arguing as in the proof of Proposition 5.3 (see e.g. (5.3)) it follows that the

polynomials in (10.36)–(10.37) have norm 1 in the corresponding norms, that is,

Ls..Pk/2/ D

Z b

apk.t/2 d� D 1;

L.b�E/.E�a/s..Pl/2/ D

Z b

apl.t/

2.b � t/.t � a/ d� D 1;

LEs�as..qk/2/ D

Z b

aqk.t/2.t � a/d� D 1;

Lbs�Es..qk/2/ D

Z b

aqk.t/

2.b� t/d� D 1:

Note that for m D 2nC 1 the polynomial PnC1 is defined (since only moments sjwith j � 2nC 1 are involved), but p

nC1 is not (because D2nC2 requires s2nC2/.Our next theorem gives explicit formulas for the function s.�/ in terms of the

orthonormal polynomials introduced above. Recall that the intervals Jj and Kl havebeen defined at the beginning of Sect. 10.6, J D [jJj and K D [lKl. As usual, Jand K denote the closures of the sets J and K, respectively. Define

m D 2n W P�.x/ D� nX

jD0pj.�/p

j.x/

�2; � 2 J; (10.38)

P�.x/ D .b � x/.x � a/

� n�1XjD0

pj.�/pj.x/

�2; � 2 K; (10.39)

m D 2nC 1 W Q�.x/ D .x � a/

� nXjD0

qj.�/q

j.x/

�2; � 2 J; (10.40)

Q�.x/ D .b � x/

� nXjD0

qj.�/qj.x/

�2; � 2 K: (10.41)

Theorem 10.29 Suppose that s 2 IntSmC1.Even Case m D 2n and � 2 J: P� is a minimizing polynomial for (10.23) and

s.�/ D� nX

jD0pj.�/2

��1:


Even Case m D 2n and � 2 K: P� is a minimizing polynomial for (10.23) and

s.�/ D 1

.b � �/.� � a/

� n�1XjD0

pj.�/2

��1:

Odd Case m D 2nC 1 and � 2 J: Q�is a minimizing polynomial for (10.23) and

s.�/ D 1

� � a

� nXjD0

qj.�/2

��1:

Odd Case m D 2nC 1 and � 2 K: Q� is a minimizing polynomial for (10.23) and

s.�/ D 1

b � �� nX

jD0qj.�/

2

��1:

Proof Since s 2 IntSmC1, Theorem 10.8 implies Dj.s/ ¤ 0 and Dj.s/ ¤ 0 forj D 0; : : : ;m. Hence the four sequences (10.32)–(10.33) are positive definite. Recallthat by Theorem 10.21 the polynomial q� from Definition 10.19 is a minimizingpolynomial of (10.23). Analyzing q� in the various cases leads to the form statedabove. We carry out the proof in the even case m D 2n; the odd case is treatedsimilarly.

Let � 2 Ji. From the list of atoms of �� given in Sect. 10.6 we know that neithera nor b are roots of �� . Therefore, by Definition 10.19, q� D g2 for some g 2 RŒx�n.Recall that q�.�/ > 0 by (10.20) and hence g.�/ ¤ 0. Setting p� WD g.�/�1gwe have p�.x/2 D q� .x/

q� .�/, so that s.�/ D Ls.q� /

q� .�/D Ls. p2�/ by Theorem 10.21. Since

s.�/ D Ls. p2�/ and p�.�/ D 1, it follows at once from the definition (10.22) of s.�/

that p� is a minimizer of Ls. p2/ for p 2 RŒx�n under the constraint p.�/ D 1. Thisproblem was settled by Proposition 9.14. By (9.34) the corresponding minimum is.Pn

jD0 pj.�/2/�1 and by (9.35) the unique minimizer is a constant multiple of the

polynomial p.x/ DPnjD0 p

j.�/p

j.x/: (Note that p

j.�/ is real, since � is real.) Hence

p2 � P� is a minimizer for (10.23). This proves the assertions for � 2 Ji.By Theorem 10.27 and the discussion preceding it, s is continuous on Œa; b�

and the continuous extension of a continuous minimizing family of polynomials for(10.23) on Ji yields minimizers for the end points of Ji. Hence the assertions remainvalid for � in the closure Ji.

Now let � 2 Ki. We proceed as in the case � 2 Ji with s replaced by thepositive definite sequence .b � E/.E � a/s: Since � 2 Ki, it follows from thedescription in Sect. 10.6 that both end points a and b are roots of �� . Therefore, byDefinition 10.19, we have q� D .b� x/.x � a/g2 with g 2 RŒx�n�1. Again g.�/ ¤ 0


by (10.20) and we set p� WD g.�/�1g. For q D .b� x/.x� a/f 2 with f 2 RŒx�n�1 wecompute

Ls.q/

q.�/D Ls..b � x/.a � x/f 2/

q.�/D Œ.b � �/.a � �/��1 L.b�E/.E�a/s. f 2/

f .�/2: (10.42)

Using Theorem 10.21, Eq. (10.42), and the relation p�.�/ D 1 we obtain

s.�/ D Ls.q�/

q�.�/D Œ.b � �/.a � �/��1 L.b�E/.E�a/s. p

2�/:

Hence it follows from the definition of s.�/ that the polynomial p� is a minimizer ofL.b�E/.E�a/s. f 2/ for f 2 RŒx�n�1 under the constraint f .�/ D 1: The orthonormalpolynomials for the sequence .b � E/.E � a/s are p0; p1; : : : ; pn�1. By Proposi-tion 9.14, applied to the sequence .b � E/.E � a/s, the corresponding minimum is.Pn�1

jD0 pj.�/2/�1 and the minimizer is a multiple of p.x/ DPn�1jD0 pj.�/pj.x/:Hence,

by (10.42), the minimum s.�/ is

Œ.b � �/.a � �/��10@n�1X

jD0pj.�/

2

1A

�1

and each multiple of .b � x/.a � x/p.x/2 � P�.x/ is a minimizer for (10.23). Thusthe assertions are proved in the case � 2 Ki. Arguing as in the preceding paragraph,the assertions hold for � in Ki as well. ut

We close this chapter by showing how the roots of principal measures �˙ andcanonical measures �� can be detected from orthogonal polynomials and quasi-orthogonal polynomials, respectively.

Proposition 10.30 The roots of the upper and lower principal representing mea-sures �C and �� of s are exactly the zeros of the following polynomials:

m D 2n �C W .b � x/Qn.x/; �� W .x � a/Qn.x/; (10.43)

m D 2nC 1 �C W .b � x/.x � a/Pn.x/; �� W PnC1.x/: (10.44)

All these zeros are simple.

Proof We carry out the proof for m D 2nC1 and �C; the other cases can be treatedsimilarly. Recall that P0;P1; : : : ;Pn are orthogonal polynomials for the sequence.b � E/.E � a/s. Hence Pn 2 RŒx�n is orthogonal to all polynomials f 2 RŒx�n�1with respect to .b � t/.t � a/d�C, that is, by (10.34) and (10.35) we have

L.b�E/.E�a/s.Pk f / DZ b

aPn.t/f .t/.b � t/.t � a/d�C D 0:


By formula (10.16),�C has n inner roots tC2 ; : : : ; tCnC1 2 .a; b/. Thus, we can choose

f to vanish at all but one such root tCi and obtain Pn.ti/.b� tCi /.tCi � a/ D 0. HencePn.ti/ D 0. Since tC1 D a and tCnC2 D b, .b � x/.x � a/Pn.x/ vanishes at all nC 2roots of �C. Because .b� x/.x� a/Pn.x/ has degree nC 2, the roots of �C exhaustthe zeros of .b � x/.x � a/Pn.x/ and all zeros are simple. ut

The roots of principal measures are described in Proposition 10.30. If � 2 Œa; b�is not a root of a principal measure, then � 2 J [ K and �� has index m C 2. InSect. 10.6 these measures have been parametrized in terms of their smallest innerroot � 2 J1 [ K1. Then �� D �� , so it suffices to know the roots of �� . The nextproposition characterizes these roots as zeros of quasi-orthogonal polynomials.

We shall say (see also Definition 9.3 below) that a polynomial q 2 RŒx�n, q ¤ 0,n � 2, is called quasi-orthogonal of order n for s if Ls.q f / D 0 for all f 2 RŒx�n�2.

Proposition 10.31 There exist strictly increasing continuous functions ' on J1 and on K1 with ranges .0;C1/ and .�1; 0/, respectively, such that for � 2 J1 [ K1the roots of the canonical measure �� are precisely the zeros of the polynomial g�defined by

m D 2n; � 2 J1 W g�.x/ D .x � a/Qn.x/� '.�/.b� x/Qn.x/; (10.45)

m D 2n; � 2 K1 W g�.x/ D .b � x/.x � a/ŒQn.x/ � .�/Qn.x/�; (10.46)

m D 2nC 1; � 2 J1 W g�.x/ D .x � a/ŒPnC1.x/� '.�/.b� x/Pn.x/�; (10.47)

m D 2nC 1; � 2 K1 W g�.x/ D .b � x/ŒPnC1.x/� .�/.x � a/Pn.x/�: (10.48)

The polynomial g� is quasi-orthogonal of order n if m D 2n and nC1 if m D 2nC1:Proof We carry out the proof in the case m D 2n and � 2 J1 D .t�1 ; t

C1 / D .a; tC1 /;

the other cases can be treated in similar manner with necessary modifications.First we fix l D 1; : : : ; nC 1 and define a function 'l on Jl by

'l.�/ D.� � a/Q

n.�/

.b � �/Qn.�/; � 2 Jl D .t�l ; tCl /: (10.49)

First we note that Qn ¤ 0 on Jl by Proposition 10.30. Hence the denominator in(10.49) is nonzero on Jl, so 'l is continuous on Jl. Since Q

nand Qn have degree n,

positive leading terms, and no zeros in .�1; a� by Proposition 10.30,Qn.�/

Qn.�/> 0 for

� D a. This holds for all � 2 J1, because Qn

and Qn do no vanish on J1. The roots

of �� and �C, hence the zeros Qn

and Qn by Proposition 10.30, strictly interlace.

Therefore,Qn.�/

Qn.�/> 0 and hence 'l.�/ > 0 on Jl for all l. Since

lim�!tCl �0

.b � �/Qn.�/ D .b � tCl /Qn.tCl / D 0 and .tCl � a/Q

n.tCl / > 0;


it follows that lim�!tCl �0 'l.�/ D C1: Similarly, lim�!t�l C0 'j.�/ D 0: Therefore,

by the continuity of the function 'l on Jl its the range is .0;C1/:We show that 'l is injective on Jl. Assume to the contrary that there exist numbers

�; � 0 2 Jl; � ¤ � 0; such that 'l.�/ D 'l.�0/. Then the polynomial g�.x/ has at least

n C 2 zeros, that is, � and � 0 in the interval Jl and one in each of the remainingn intervals Ji. But deg.g�/ � n C 1, so that g� D 0. Setting x D b we get acontradiction, since b is not a root of �� and hence g�.b/ D .b � a/Q

n.b/ ¤ 0

by Proposition 10.30. Summarizing, we have shown that 'l is a strictly increasingcontinuous function on Jl with range .0;C1/.

Now we set ' WD '1. Recall that Qn

and Qn are orthogonal polynomials withrespect to .x � a/d�� and .b � x/d�� , respectively. Hence, for f 2 RŒx�n�1,

Z b

ag� f d�� D

Z b

aQ

n.x/f .x/.x � a/ d�� '.�/

Z b

aQn.x/f .x/ .b � x/ d�� D 0:

(10.50)

The measure �� has nC 1 roots tl.�/ 2 Jl; l D 1; : : : ; nC 1, where t1.�/ D �. By(10.49), we have g�.t1.�// D g�.�/ D 0: Thus, if we take a polynomial f 2 RŒx�n�1that vanishes at all t j.�/, j D 2; : : : ; nC 1; except ti.�/, it follows from (10.50) thatg� vanishes at ti.�/. Hence all roots of �� are zeros of g� . Since deg.g�/ � n C 1,these nC 1 roots exhaust the zeros of g� .

Finally, we show that g� is quasi-orthogonal. As a sample we verify this in thecase m D 2n, � 2 K1. Let f 2 RŒx�n�2. Using again the orthogonality of Qn and Q

nwe derive

Ls.g� f / D Ls..b � x/.a � x/ŒQn.x/ � .�/ Qn.x/� f /

D Ls..x � a/Qn.x/.b � x/f / � .�/ Ls..b� x/Qn.x/.x � a/f /

D Lbs�Es.Qn.x/.x � a/f /� .�/ LEs�as.Qn.x/.b � x/f / D 0;

where the last equality follows from (10.34), since .x � a/f ; .b � x/f 2 RŒx�n�1:Hence g� is a quasi-orthogonal polynomial of order n. ut

10.8 Exercises

1. Let Œa; b� D Œ0; 1� and s D .1; s1/ 2 S2. Draw a picture of S2 and compute thenumbers sC

2 and s�2 .

2. Let Œa; b� D Œ�1; 1�, m D 2 and s D .1; 0; 0/. Compute the canonical measure�� for � 2 Œa; b� and determine the principal measures �C and ��.

3. Suppose that a � 0 and b � 2. Let m D 4 and s D .8; 6; 12; 24; 48/:a. Show that s 2 S5.b. Show that s is determinate and compute the unique representing measure of s.

10.9 Notes 255

c. Is s0 D .8; 6; 12; 24/ 2 S4 determinate?d. Is s00 D .8; 6; 12; 24; 48; 96/ 2 S6 determinate?

4. Let c > 0; d > 0. Set Œa; b� D Œ�c; c�, m D 3 and s D .2dC 2; 0; 2c2; 0/:a. Show that s 2 IntS4.b. Prove that the two principal measures �C and �� are given by

�C D ı�c C 2dı0 C ıc; �� D .dC 1/ı�� C .dC 1/ı� ; where � WD c.d C 1/�1=2:

5. Let � D PkjD1mjıtj be a canonical representing measure for s 2 IntSmC1 and

let P 2 RŒx�m be a polynomial which has simple zeros at t1; : : : ; tk. Prove that

mj D Ls

�P.x/

P0.tj/.x � tj/

�for j D 1; : : : ; k:

6. Determine the extreme rays of the moment cone SmC1:

10.9 Notes

According to M.G. Krein [Kr2], the ideas of this chapter go back to the Russianmathematicians P.L. Tchebycheff and A.A. Markov. Markov invented canonicalmeasures and applied them in his study of “limiting values” of integrals [Mv2]. Thetheory of principal measures, canonical measures, and maximal masses presentedabove is taken from the fundamental paper [Kr2].

The geometry of moment spaces SmC1 was elaborated by S. Karlin and L.S.Shapley [KSh]. The volume of the projection of the base Sm of the moment cone inRm is computed in [KSh]. Theorem 10.29 is due to I.J. Schoenberg and G. Szegö[SSz], improving a result of Krein. The two classical monographs of S. Karlin andW.J. Studden [KSt] and of M.G. Krein and A.A. Nudelman [KN] contain furtherresults and a detailed study of Tchebyscheff systems, see also [DS]. We partlyfollowed these books. A description of all solutions of various types of truncatedmoment problems is given by Krein [Kr3].

Chapter 11The Moment Problem on the Unit Circle

This chapter is concerned with the trigonometric moment problem:Let s D .sj/j2N0 be a complex sequence. When does there exist a Radon measure

� on the unit circle T such that for all j 2 N0,

sj DZT

z�jd�.z/‹ (11.1)

The truncated trigonometric moment problem is the corresponding problem for afinite sequence .sj/njD0 of prescribed moments. The aim of this chapter is to a give acondensed treatment of some basic notions and results on these problems.

In Sect. 11.1 we prove the Fejér–Riesz theorem (Theorem 11.1) on nonnegativeLaurent polynomials on T. This is the key result for solving the trigonometricmoment problem (Theorem 11.3) in Sect. 11.2. Section 11.3 deals with orthogonalpolynomials on the unit circle. The Szegö recurrence relations (Theorem 11.9) andVerblunsky’s theorem (Theorem 11.12) about the reflection coefficients occuringin these relations are obtained. In Sect. 11.4 the truncated trigonometric momentproblem is investigated. In Sect. 11.5 we give a short digression into Carathéodoryand Schur functions and the Schur algorithm and prove Geronimus’ theorem(Theorem 11.31) about the equality of reflection coefficients and Schur parameters.

Throughout this chapter we adopt the following notational convention which isoften used without mention: For a sequence .sj/njD0, where n 2 N or n D1, we sets�j WD sj for j � 1. The reason is that (11.1) holds for �j provided it does for j.

11.1 The Fejér–Riesz Theorem

The solution of the moment problem on the unit circle is essentially based on thefollowing Fejér–Riesz theorem.


257

258 11 The Moment Problem on the Unit Circle

Theorem 11.1 Suppose that p.z/ D PnkD�n akz

k 2 CŒz; z�1�, p ¤ 0, is a Laurentpolynomial such that p.z/ is real and nonnegative for all z 2 T.

Then there exists a unique polynomial q.z/ D PnjD0 bkzk 2 CŒz� of the same

degree such that q.z/ ¤ 0 for jzj < 1, q.0/ > 0, and

p.z/ D jq.z/j2 for z 2 T: (11.2)

Proof Without loss of generality we can assume that n 2 N and an ¤ 0. Since p.z/is real on T, we have a�k D ak for all k. Put f .z/ WD znp.z/. Then f 2 CŒz� hasdegree 2n , f .0/ D an ¤ 0, and the nonzero zeros of f and p coincide. Clearly,

f .z/ D anz2n C � � � C a0z

n C � � � C an D z2n f .z�1/; z 2 C; z ¤ 0: (11.3)

Equation (11.3) implies that the zeros of f are symmetric with respect to the unitcircle. More precisely, if w is a zero of f , then w ¤ 0 and w�1 is also a zero of fwith the same multiplicity. Let z1; z�1

1 ; : : : ; zm; z�1m denote the zeros of f which are

not on T (if there are such zeros) counted according to their multiplicities.Define g.x/ D p.eix/ for x 2 R. By differentiation it follows that each zero ei� 2

T of p has the same multiplicity as the zero � 2 R of g. Since g.x/ D p.eix/ � 0 onR, each zero � of g, hence each zero ei� 2 T of p and so of f , is of even multiplicity.If f has zeros on T, we denote them by �1; �1; : : : ; �l; �l, so that 2mC 2l D 2n.

Then the polynomial f and hence p factor as

p.z/ D z�nf .z/ D z�nan

mYkD1.z� zk/.z � z�1

k /

lYjD1.z � �j/2

DmY

kD1.z� zk/.z

�1 � zk/lY

jD1.z � �j/.z�1 � � j/

z�nan

mYkD1.�z/z�1

k

lYjD1.�z/�j

:

The factor in square brackets is czmCl�n for some c 2 C. Since mC l D n and p � 0on T, it follows that czmCl�n D c > 0. Setting

q0.z/ Dpc

mYkD1.z� zk/

lYjD1.z � �j/;

the preceding equality yields p.z/ D jq0.z/j2 for z 2 T. (One of the two groups ofzeros may be absent. In this case we set the corresponding product equal to one.)

Upon multiplying q0 by a constant of modulus one, we can have q0.0/ > 0.

Recall that z 7! 1�� zz�� is a bijection of T for any � 2 C. Therefore, if � is a zero of

q0, the polynomial q0.z/1�� zz�� has the same degree as q0 and satisfies (11.2) as well.

Continuing in this manner, we can remove all zeros of q0 which are contained inD D fz 2 C W jzj < 1g and obtain a polynomial q which has the desired properties.

11.2 Trigonometric Moment Problem: Existence of a Solution 259

Let Qq be another polynomial with these properties. By (11.2), jq.z/j D Qq.z/jon T: From this it follows that the zeros of q and Qq on T and their multiplicitiescoincide. Since q.z/ ¤ 0 and Qq.z/ ¤ 0 on D, the functions f .z/ WD Qq.z/

q.z/ and 1f .z/ are

holomorphic on D and satisfy j f .z/j D j 1f .z/ j D 1 on T. By the maximum principle

for holomorphic functions, j f .z/j � 1 and j 1f .z/ j � 1, hence j f .z/j D 1, on D.Since j f .z/j attains its maximum in D, f is constant. From q.0/ > 0, Qq.0/ > 0; andj f .0/j D 1 we get f .0/ D 1. Hence f .z/ � 1, so that q D Qq. utRemark 11.2 The polynomial q in Theorem 11.1 satisfies

p.z/ D q.z/ q.z�1/ for z 2 C; z ¤ 0: ı

11.2 Trigonometric Moment Problem: Existenceof a Solution

Recall that the group �-algebra CŒZ� of Z is the unital �-algebra CŒz; z�1� of allLaurent polynomials p.z/ D Pn

jD�n cjzj, where cj 2 C and n 2 N, with involution

p 7! p�.z/ D PnjD�n cjz

�j. In the �-algebra CŒZ� we have z� D z�1, that is, z is aunitary element. The character space of CŒZ� is the torus T D fz 2 C W jzj D 1g,where z 2 T acts on CŒZ� by the point evaluation �z. p/ D p.z/:

Note that CŒZ� can be considered as the �-algebra of trigonometric polynomials

f .�/ DnX

jD�n

cjeij� D c0 C

nXlD1.al cos l� C bl sin l�/; � 2 Œ��; ��;

where cj; al; bl 2 C and al D cl C c�l; bl D i.cl � c�l/, with involution given byf 7! f �.�/ DPn

jD�n cje�ij� :

Let s D .sj/j2N0 be a sequence. We denote by Ls the linear functional on CŒZ�

given by Ls.z�j/ WD sj; j 2 Z. Recall from Example 2.3.3 that the Hankel matrix isnow the infinite Toeplitz matrix H.s/ D .hjk/j;k2N0 with entries hjk WD sk�j; j; k 2N0: Here we have set s�l D sl for l � 1 according to our notational convention.

If s D .sj/mjD0 is a sequence and n � m, then Hn.s/ D .hjk/nj;kD0 denotes the.nC1/ � .nC1/ Toeplitz matrix with entries hjk D sk�j and we abbreviate

Dn WD Dn.s/ � detHn.s/ D

ˇˇˇˇˇˇ

s0 s1 : : : sns�1 s0 : : : sn�1s�2 s�1 : : : sn�2: : : : : : : : : : : :

s�n s�nC1 : : : s0

ˇˇˇˇˇˇ

(11.4)


The next theorem, called the Carathéodory–Toeplitz theorem, provides a solutionof the trigonometric moment problem.

Theorem 11.3 For a complex sequence s D .sn/n2N0 the following areequivalent:

(i) s is a moment sequence for the group Z, that is, there exists a Radon measure� on T such that

sn DZT

z�nd�.z/ for all n 2 Z: (11.5)

(ii) Ls.q�q/ � 0 for q 2 CŒZ�, i.e. Ls is a positive functional on the �-algebraCŒZ�.

(iii) Ls.q�q/ � 0 for all q 2 CŒz�.(iv) The infinite Toeplitz matrix H.s/ D .sk�j/

1j;kD0 is positive semidefinite.

(v)P1

j;kD0 sj�kck cj � 0 for all finite complex sequences .cj/j2N0 .

The measure � is uniquely determined by (11.5). Its support is an infinite set ifand only if the Toeplitz matrix H.s/ is positive definite, or equivalently, the momentsequence s is positive definite, or equivalently, Dn.s/ > 0 for all n 2 N0.

Proof Let q.z/ DPnkD0 ckzk 2 CŒz�. Then q�.z/ DPn

jD0 cj z�j and

Ls.q�q/ D

nXj;kD0

Ls.zk�j/ck cj D

1Xj;kD0

sj�k ck cj D1X

j;kD0hkj ckcj: (11.6)

From (11.6) we conclude that (iii)$(iv)$(v). Further, (i)!(ii) by Proposition 2.7and (ii)!(iii) is trivial. We prove the implication (iii)! (i).

Let p 2 CŒZ� be such that p.z/ � 0 on T. By the Fejér–Riesz theorem 11.1, thereis a polynomial q DPn

jD0 cjzj 2 CŒz� such that p D q�q. Then Ls. p/ D Ls.q�q/ �0 by (iii). Hence the restriction of Ls to the real subspace E WD f p 2 CŒZ� W p D p�gof C.TIR/ is EC-positive. Therefore, by Proposition 1.9, the restriction of Ls to E,hence also Ls on CŒZ�, is given by a measure � 2 MC.T/. This implies (i).

Thus we have shown that the four conditions (i)–(iv) are equivalent.Since the trigonometric polynomials are dense in C.T/ by Fejér’s theorem, the

measure � is uniquely determined by (i).We verify the last assertion. If q.z/ is as above, from (11.6) and (i) we obtain

nXj;kD0

hkj ck cj DnX

j;kD0ck cj

ZT

zk�jd� DZT

ˇˇ nXkD0

ckzk

ˇˇ2d� D

ZT

jq.z/j2d�:

(11.7)

If � has finite support, we can find q ¤ 0, which vanishes on supp�. Then.c0; : : : ; cn/ ¤ 0 and

Pnj;kD0 hkj ckcj D 0 by (11.7). That is, the matrix H.s/ and the

sequence s are not positive definite.

11.3 Orthogonal Polynomials on the Unit Circle 261

If � has infinite support, for any vector .c0; : : : ; cn/ ¤ 0 the polynomial q doesnot vanish on supp�. Hence it follows from (11.7) that

Pnj;kD0 hkj ck cj > 0, that is,

H.s/ and s are positive definite.Obviously, the infinite matrix H.s/ is positive definite if and only if all finite

matrices Hn.s/ are, or equivalently, Dn.s/ > 0 for all n 2 N0: utRemark 11.4

1. The minus signs in (11.5) and in the definition Ls.z�j/ D sj are notationalconventions following the standard literature [KN], [Sim2]. The minus sign in(11.5) also fits into the usual definition of the Fourier transform for the group Z.It should be noted that some authors define Ls.zj/ D sj and/or hjk D sk�j:

2. In (i), we have set s�n WD sn for n 2 N by our convention. In (iii), the conditionLs.q�q/ � 0 is required only for “analytic” polynomials q.z/DPn

jD0 cjzj. ıThe following theorem is the counterpart of Theorem 11.3 for the truncated

trigonometric moment problem.

Theorem 11.5 Let n 2 N0. For a sequence.sj/njD0 the following are equivalent:

(i) There is a Radon measure � on T such that

sj DZ

z�jd�.z/ for j D 0; : : : ; n: (11.8)

(ii) There is a k-atomic measure � on T, k � 2nC 1, such that (11.8) holds.(iii) The Toeplitz matrix Hn.s/ is positive semidefinite.(iv)

Pnj;kD0 sj�kck cj � 0 for all .c0; : : : ; cn/T 2 CnC1.

Proof (ii)!(i) is trivial. Similarly, as in the proof of Theorem 11.3 the implications(i)!(iii)$(iv) follow from (11.6). It suffices to prove that (iv) implies (ii). Let E bethe real vector space of polynomials p D p� 2 CŒZ� such that deg. p/ � n: Clearly,dimE D 2nC 1. Let p 2 E be such that p.z/ � 0 on T. Then the polynomial q fromthe Fejér–Riesz theorem also satisfies deg q � n. Hence it follows from (11.6) and(iv) that Ls. p/ D Ls.q�q/ � 0. Thus Proposition 1.26 applies and yields (ii). ut

11.3 Orthogonal Polynomials on the Unit Circle

In this section, we suppose that s D .sj/j2Z is a moment sequence on Z such that itsrepresenting measure � on T has infinite support. By Theorem 11.3 the latter holdsif and only if the infinite Toeplitz matrix H.s/ is positive definite.

Since H.s/ is positive definite, there is a scalar product h�; �is on CŒz� given by

h p; qis WDnX

j;kD0ajbk sk�j (11.9)


for p D PnjD0 ajzj 2 CŒz� and q D Pn

kD0 bkzk 2 CŒz�. From the definition (11.9) itis immediate that sk D h1; zkis for k 2 Z and

hzp; zqis D h p; qis: (11.10)

Next we define two families of polynomials Pk, P�k associated with s. Put

D�1 WD 1 and P0.z/ WD 1. For k 2 N we set

Pk.z/ WD 1

Dk�1

ˇˇˇˇˇˇ

s0 s1 : : : sk�1 1s�1 s0 : : : sk�2 zs�2 s�1 : : : sk�3 z2: : : : : : : : : : : : : : :

s�k s�kC1 : : : s�1 zk

ˇˇˇˇˇˇ: (11.11)

Since the coefficient of zk in the determinant is Dk�1, Pk is a monic polynomialof degree k for k 2 N0. The next lemma shows that Pk; k 2 N0, are orthogonalpolynomials of the unitary space .CŒz�; h�; �is/.Lemma 11.6 hPk;Pjis D D�1

k�1Dkıkj and hPk; zkis D D�1k�1Dk for k; j 2 N0.

Proof Let l D 0; : : : ; k: To compute hPk.z/; zlis we repeat the reasoning from theproof of Proposition 5.3. Multiplying the last column by z�l and applying the func-tional Ls the last column of the determinant will be replaced by .sl; sj�1; : : : ; sj�k/

T .If l < k, the last column coincides with the j-th column and hence

hPk.z/; zlis D 0: Since Pk has degree k, this implies that hPk;Pjis D 0 for allk ¤ j; k; j 2 N0.

Now let j D k. If j D k D 0, then obviously hP0;P0is D hP0; z0is D s0 D D0.For j D k 2 N, the determinant becomes Dk, so that hPk; zkis D D�1

k�1Dk: Hence,since the leading coefficient of Pk is 1, we get hPk;Pkis D hPk; zkis D D�1

k�1Dk: utLet p DPk

jD0 cjzj be a polynomial of degree at most k. Put p WDPkjD0 cjzj. The

reciprocal polynomial Rk. p/ is defined by

Rk. p/.z/ WD zkp.z�1/ DkX

jD0ck�jz

j:

Clearly, deg.Rk. p// � k and deg.Rk. p// D k if and only if p.0/ ¤ 0. Then we have

.Rk. p//.z/ D zk p.z/ D zk p.1=z/ for z 2 T; (11.12)

hRk. p/;Rk.q/is D hq; pis: (11.13)

For p D Pk we abbreviate P�k WD Rk.Pk/. Since Pk is monic, P�

k .0/ D 1. Clearly,

P0.z/ D P�0 .z/ D 1; P1.z/ D z � s�1s�1

0 ; P�1 .z/ D 1� s1s

�10 z:


Combining the definition P�k .z/ D Rk.Pk/.z/ D zkPk.z�1/ with (11.11) we obtain

the explicit formula

P�k .z/ D

1

Dk�1

ˇˇˇˇˇˇ

s0 s�1 : : : s�kC1 zks1 s0 : : : s�kC2 zk�1s2 s1 : : : s�kC3 zk�2: : : : : : : : : : : : : : :

sk sk�1 : : : s1 1

ˇˇˇˇˇˇ; k 2 N: (11.14)

Remark 11.7 The notation P�k is standard in the literature. Note that P�

k should notbe confused with the adjoint of Pk in the �-algebra CŒZ�! ı

Some simple facts on these polynomialsP�k are collected in the following lemma.

Lemma 11.8

(i) hP�k ; z

jis D 0 for j D 1; : : : ; k and hP�k ; 1is D D�1

k�1Dk.(ii) kPkk2s D kP�

k k2s D hP�k ; 1is D D�1

k�1Dk for k 2 N0.

Proof All assertions follow by a repeated application of Lemma 11.6 combinedwith (11.13). For j D 1; : : : ; k we obtain

hP�k ; z

jis D hRk.Pk/;Rk.zk�j/is D hzk�j;Pkis D 0:

Further, again by (11.13), we have kP�k k2 D kPkk2 and

hP�k ; 1i D hRk.Pk/;Rk.z

k/i D hzk;Pki D kPkk2 D D�1k�1Dk: ut

The following theorem is the first main result of this section. The formulas(11.15) and (11.16) therein are called Szegö recursion formulas.

Theorem 11.9 Suppose that s D .sj/j2Z is a positive definite sequence on Z. Thenthere exist uniquely determined complex numbers ˛n for n 2 N0 such that

PnC1.z/ D zPn.z/ � ˛nP�n .z/; (11.15)

P�nC1.z/ D P�

n .z/ � ˛nzPn.z/: (11.16)

Further, we have

˛n D �PnC1.0/ ; (11.17)

kPnC1k2s D .1 � j˛nj2/kPnk2s D s0

nYjD0.1 � j˛jj2/; (11.18)

DnC1 D Dn s0

nYjD0.1 � j˛jj2/: (11.19)


Proof Using Lemma 11.6 and formula (11.10) we obtain for j D 1; : : : ; n;

hPnC1 � zPn; zji D hPnC1; z ji � hPn; z

j�1i D 0:

By Lemma 11.8, P�n is also orthogonal to z; z2; : : : ; zn. Since PnC1 and Pn are monic,

PnC1� zPn has degree at most n. Thus PnC1� zPn and P�n are both of degree at most

n and orthogonal to z; : : : ; zn. Therefore, since hP�n ; 1i ¤ 0 by Lemma 11.8, setting

˛n WD hP�n ; 1i�1s hPnC1 � zPn; 1is we conclude that PnC1 D zPn � ˛nP�

n which is(11.15). Applying RnC1 to both sides of (11.15) yields (11.16).

Setting z D 0 in (11.15) and using that P�n .0/ D 1 we obtain (11.17). In

particular, this implies that ˛n is uniquely determined by (11.15).Next we prove (11.18). Recall that multiplication by z is unitary by (11.10),

PnC1?P�n (since deg.P�

n / � n) and kPnk D kP�nk by Lemma 11.8(ii). Using these

facts it follows from (11.15) that

kPnk2s D kzPnk2s D kPnC1 C ˛nP�nk2s D kPnC1k2s C j˛nj2kPnk2s ;

which implies the first equality of (11.18). The second follows by repeatedapplication of the first combined with fact that kP0k2s D h1; 1is D s0.

Inserting the equality kPnC1k2s D D�1n DnC1 (from Lemma 11.8(ii)) into (11.18)

we obtain (11.19). utDefinition 11.10 The numbers ˛n from (11.15) are called the reflection coefficientsof s; they are also denoted by ˛n.s/, or ˛n.�/, where � is the unique representingmeasure of s.

Remark 11.11

1. The numbers ˛n also appear under the names Verblunsky coefficients in [Sim2],canonical moments in [DS], or Schur parameters in the literature.

2. The choice of writing �˛n in (11.15) follows [Sim2]. The reason is that then˛n.�/ becomes equal to the Schur parameter �n.�/ by Geronimus’ theorem11.31 below.

3. Equations (11.15) and (11.16) can be rewritten in matrix form as

�PnC1.z/P�nC1.z/

�D�

z �˛n

�˛nz 1

��Pn.z/P�n .z/

�: ı

The second main result of this section is the following Verblunsky theorem. Itstates that sequences of numbers ˛n 2 D are precisely the sequences of possiblereflection coefficients of probability measures on T of infinite support. Sincethe parameters ˛n appearing in the Szegö relation (11.15) are the counterpartof the Jacobi parameters an; bn from Sect. 5.2, Theorem 11.12 might be called“Favard’s theorem for the unit circle”.


Theorem 11.12 For a complex sequence .˛n/1nD0 the following are equivalent:

(i) There is a positive definite sequence s D .sj/j2Z, where s0 D 1, on Z such that˛n D ˛n.s/ for n 2 N0.

(ii) There exists a probability measure � on T of infinite support such that ˛n D˛n.�/ for n 2 N0.

(iii) ˛n 2 D for all n 2 N0.

By Theorem 11.3, a moment sequence on Z is positive definite if and only ifits representing measure has infinite support. Hence Theorem 11.3 yields (i)$(ii).(i)!(iii) follows at once from formula (11.18), since PnC1 ¤ 0 and (11.18) implythat j˛nj < 1. The main implication (iii)!(i) will be proved at the end of the nextsection.

We close this section by stating some facts on zeros of the polynomials Pn, P�n .

Proposition 11.13 For n 2 N we have:

(i) If z0 2 C is a zero of Pn.z/, then jz0j < 1.(ii) If z0 2 C is a zero of P�

n .z/, then jz0j > 1:(iii) jP�

n .z/j D jPn.z/j for z 2 T.(iv) jP�

n .z/j < jPn.z/j for z 2 D.

Proof

(i) Since Pn.z0/ D 0, p.z/ WD Pn.z/z�z0

is a polynomial of degree n � 1. Hence Pn isorthogonal to z0p in the unitary space .CŒz�; h�; �is/: Using this fact we derive

kpk2s D kzpk2s D k.z � z0/pC z0pk2s D kPn C z0pk2s D kPnk2s C jz0j2kpk2s ;

so that

.1 � jz0j2/kpk2s D kPnk2s : (11.20)

Since h�; �is is a scalar product, we have kpks > 0 and kPnks > 0. Therefore(11.20) implies that jz0j < 1:

(ii) Recall that P�n .z/ D Rn.Pn/.z/ D znPn.z�1/ and P�

n .0/ D 1. Hence P�n .z0/ D 0

implies z0 ¤ 0 and Pn.z�10 / D 0, so that jz0j > 1 by (i).

(iii) follows from jP�n .z/j D jznPn.z/j D jPn.z/j for z 2 T by (11.12).

(iv) By (ii), P�n .z/ ¤ 0 for jzj � 1. Hence the function f .z/ WD Pn.z/

P�

n .z/is holomorphic

on D, continuous on the closure of D, and of modulus one on T D @D by(ii). Since n > 0, Pn.z/ has a zero in D, so in particular, f .z/ is not constant.Therefore, by the maximum principle for holomorphic functions, j f .z/j < 1

and hence jP�n .z/j < jPn.z/j for z 2 D. ut


11.4 The Truncated Trigonometric Moment Problem

Definition 11.14 Let n 2 N. The moment cone SnC1 is given by

SnC1 WD�s D .s0; s1; : : : ; sn/ W sj D

ZT

z�j d�.z/; j D 0; : : : ; n; � 2 MC.T/�:

As in Sect. 10.2, we consider SnC1 as a subset of CnC1 by identifying s with thecolumn vector sT 2 CnC1. By a similar reasoning as in the proof to Proposition 10.5it follows that SnC1 is a closed convex cone in CnC1 and the conic convex hull ofthe moment curve

cnC1 D fs.z/ D .1; z; z2; : : : ; zn/ W z 2 Tg:

The latter implies that each s 2 SnC1 has an atomic representing measure

� DkX

jD1mjızj ; (11.21)

where z1; : : : ; zk are pairwise different points of T and mj > 0 for j D 1; : : : ; k. Forsuch a measure � we set ind.�/ D 2k. (Since the unit circle T has no end points,each atom of � is counted twice.) The index ind.s/ is defined as the minimum ofindices of all such representing measures (11.21) for s.

Let s D .sj/njD0 2 SnC1. By our notational convention, we have defined s�j WD sjfor j D 1; : : : ; n. Hence, if � is a representing measure for s, then

sj DZT

z�j d� for j D �n;�nC 1; : : : ; n:

By a slight abuse of notation we denote the “double” sequence .sj/njD�n also by s.Recall that by Theorem 11.5 a complex sequence s D .sj/njD0 belongs to SnC1 if

and only if the Toeplitz matrix Hn.s/ is positive semidefinite.The following propositions characterizes boundary points and interior points of

the set SnC1. The proofs are verbatim the same as for its counterparts on a boundedinterval (Theorems 10.7 and 10.8) and will be omitted.

Proposition 11.15 A sequence s 2 SnC1 is a boundary point of SnC1 if and onlyif ind.s/ � 2n. In this case s is determinate, that is, it has a unique representingmeasure � 2 MC.T/.

Proposition 11.16 For s 2 SnC1 the following statements are equivalent:

(i) s 2 IntSnC1, that is, s is an interior point of SnC1.(ii) The Toeplitz matrix Hn.s/ is positive definite.

(iii) Dj.s/ > 0 for j D 0; : : : ; n.

11.4 The Truncated Trigonometric Moment Problem 267

Note in conditions (ii) and (iii) of Proposition 11.16 the numbers s�j D sj for j D1; : : : ; n are required.

Now we fix s D .sj/njD0 2 IntSnC1. Then, by Proposition 11.16, Dj.s/ > 0 forj D 0; : : : ; n and hence the sequence s D .sj/njD�n is positive definite, that is,

nXj;kD0

sj�kck cj > 0 for all .c0; : : : ; cn/T 2 CnC1; .c0; : : : ; cn/ ¤ 0:

This implies that Eq. (11.9) defines a scalar product on the vector space CnŒz�.Further, proceeding as in the last section, we define the polynomials Pk and P�

k ,k D 0; : : : ; n; and the reflection coefficients ˛j; j D 0; : : : ; n � 1, and derive thecorresponding properties from Theorem 11.9.

Let us denote byCs the set of numbers snC1 2 C for which the extended sequenceQs WD .s0; : : : ; sn; snC1/ belongs to SnC2: In the proof of Proposition 11.17 below weshall use the following notation:

�nC1.z/ D

ˇˇˇˇ

s0 s1 : : : sn zs�1 s0 : : : sn�1 sn: : : : : : : : : : : : : : :

z s�n : : : s�1 s0

ˇˇˇˇ ; (11.22)

An.z/ D

ˇˇˇˇ

s1 s2 : : : sn zs0 s1 : : : sn�1 sn: : : : : : : : : : : :

s�nC1 s�nC2 : : : s0 s1

ˇˇˇˇ ; (11.23)

Bn.z/ D

ˇˇˇˇ

s�1 s0 : : : sn�2 sn�1s�2 s�1 : : : sn�3 sn�2: : : : : : : : : : : :

z s�nC1 : : : s�2 s�1

ˇˇˇˇ : (11.24)

The next proposition is the counterpart of Corollary 10.16 for the unit circle.

Proposition 11.17 Suppose that s D .sj/njD0 2 IntSnC1. Then the set Cs is a

closed disk with radius rnC1 D Dn.s/Dn�1.s/

and center cnC1 D .�1/nC1An.0/

Dn�1.s/.

Proof By Theorem 11.5 (i)$(iii), Cs is the set of complex numbers snC1 suchthat HnC1.Qs/ is positive semidefinite. Since Dj.s/ > 0 for D 1; : : : ; n by Proposi-tion 11.16, Cs is precisely the set of numbers z D snC1 2 C for which�nC1.z/ � 0:


To describe this set we apply Sylvester’s formula (see e.g. [Gn, p. 58]) andexpand the determinant �nC1.z/ with respect to the first and last row and first andlast column. Then we obtain

Dn�1.s/�nC1.z/ D Dn.s/2 � An.z/Bn.z/: (11.25)

Since sj D s�j, we have Bn.z/ D An.z/. Clearly, the determinant An.z/ is a linearpolynomial in z with leading coefficient .�1/nDn�1.s/. Therefore,

An.z/ D z.�1/nDn�1.s/C An.0/: (11.26)

Hence, since Dn�1.s/ > 0, it follows from (11.25) that �nC1.z/ � 0 if and only if

Dn.s/2 � jAn.z/j2 D jz.�1/nDn�1.s/C An.0/j2;

or equivalently,

rnC1 � Dn.s/

Dn�1.s/�ˇˇz� .�1/nC1 An.0/

Dn�1.s/

ˇˇ � jz� cnC1j: (11.27)

This completes the proof of Proposition 11.17. utThe closed disk Cs from Proposition 11.17 is the set of the possible .n C 1/-th

moments snC1, or more precisely, the set of numbers snC1 for which the extendedsequence Qs D .s0; : : : ; sn; snC1/ is in SnC2: Recall that DnC1.Qs/ D �.snC1/by construction and Dj.s/ D Dj.Qs/ > 0 for j D 0; : : : ; n by the assumptions 2 IntSnC1. Therefore, by Proposition 11.15, Qs belongs to boundary of MnC2if and only if �.snC1/ D 0, or equivalently, snC1 lies on the circle @Cs.

For � 2 @Cs let PnC1.zI �/ denote the polynomial (11.11) with k D n C 1 ands�.nC1/ D � . All other moments sj required in (11.11) are determined by s D .sj/njD0:

Let us call a representing measure � for s canonical if ind.�/ D 2nC 2.

Theorem 11.18 Suppose that s D .sj/njD0 2 IntSnC1: For each � on the circle@Cs there exists a unique canonical representing measure �� for s such that

� D snC1.��/ DZT

z�.nC1/d��.z/: (11.28)

The .n C 1/ atoms of �� are precisely the roots of the polynomial PnC1.zI �/. Inparticular, PnC1.zI �/ has nC 1 distinct simple roots, all of them lying on T.

Proof Let Qs D .s0; : : : ; sn; snC1/, where snC1 WD �. Because � 2 @Cs, we haveDnC1.Qs/ D �nC1.�/ D 0, so Qs is a boundary point of SnC2. Therefore, byProposition 11.15, Qs has a unique representing measure�� . Clearly,�� is the uniquerepresenting measure for s which satisfies (11.28).

11.4 The Truncated Trigonometric Moment Problem 269

Since Qs belongs to the boundary of SnC2, ind.��/ � 2n C 2 by Proposi-tion 11.15. Further, ind.��/ � 2n would imply that s is in the boundary of SnC1,which contradicts the assumption s 2 IntSnC1: Thus ind.��/ D 2nC 2 and ��is indeed a canonical measure for s.

Since HnC1.Qs/ is positive semidefinite, Eq. (11.9), with s replaced by Qs, defines anonnegative sesquilinear form h�; �iQs on CnC1Œz�. Repeating the proof of Lemma 11.6and using that �� is a representing measure of Qs we derive

ZT

jPnC1.zI �/j2d��.z/ D hPnC1;PnC1iQs D Dn.s/�1DnC1.Qs/ D 0:

Therefore, all nC1 atoms of �� are zeros of PnC1.zI �/ (by Proposition 1.23). SinceDn.s/ > 0 and hence deg.PnC1/ D n C 1, these atoms exhaust the zeros of PnC1and all zeros are simple. The atoms of �� are in T. Hence the zeros of PnC1 arein T. ut

Next we consider extensions Qs contained in IntSnC2. Our aim is to build thebridge to the reflection coefficients. Since we assumed that s 2 IntSnC1, thesequence s is positive definite and hence (11.9) defines a scalar product on the vectorspace CnŒz�. Then, proceeding as in the last section, the orthogonal polynomialsPk; k D 0; : : : ; n; and the reflection coefficients ˛j; j D 0; : : : ; n� 1; are defined andthe corresponding properties from Theorem 11.9 remain valid. By (11.19),

rnC1 D Dn�1.s/�1Dn.s/ D s0

n�1YjD0.1 � j˛jj2/:

Obviously, z D snC1 is in the interior of Cs if and only if jsnC1 � cnC1j < rnC1:Suppose that ˛n 2 D WD fz 2 C W Œzj < 1g is given. Then, setting

snC1 WD cnC1 C ˛nrnC1 � .�1/nC1An.0/

Dn�1.s/C ˛n s0

n�1YjD0.1 � j˛jj2/; (11.29)

snC1 belongs to the interior of Cs. Then DnC1.Qs/ D �.snC1/ > 0, so Qs D .sj/nC1jD0

belongs to Int SnC2 by Proposition 11.16. Conversely, if Qs 2 IntSnC2, then snC1is in the interior of Cs and hence snC1 is of the form (11.29) for some unique ˛n 2 D.

Formula (11.29) describes the new moment snC1 in terms of the given number ˛nand of the reflection coefficients ˛j and moments sj, j < n, of s.

The following formula expresses ˛n in terms of the moments sj; j D 0; : : : ; nC1:

˛n D .�1/nDn.s/

ˇˇˇˇ

s1 s2 : : : sn snC1s0 s1 : : : sn�1 sn: : : : : : : : : : : :

s�nC1 s�nC2 : : : s0 s1

ˇˇˇˇ : (11.30)


We prove formula (11.30). Indeed, first using (11.29), then the formulas for rnC1and cnC1, and finally formula (11.26) we derive

˛n D r�1nC1ŒsnC1 � cnC1� D Dn�1.s/

Dn.s/

�snC1 � .�1/nC1An.0/Dn�1.s/�1

�

D .�1/nDn.s/

�snC1.�1/nDn�1.s/C An.0/

� D .�1/nDn.s/

An.snC1/:

Inserting the definition (11.23) of An.snC1/ into the right-hand side we obtain(11.30). This completes the proof of formula (11.30).

Finally, we prove that ˛n is the n-th reflection coefficient of the sequence Qs. Leten denote the n-th reflection coefficient of Qs: If PnC1 is the .n C 1/-th orthogonalpolynomial for Qs, then en D �PnC1.0/ by (11.17). We develop the determinant informula (11.11) for PnC1.0/ by the last column and apply the complex conjugate.Then �PnC1.0/ becomes the right-hand side of (11.30). Hence en D ˛n, that is, ˛nis the n-th reflection coefficient of Qs.

We summarize the preceding considerations in the following theorem.

Theorem 11.19 Suppose that s D .sj/njD0 2 IntSnC1. There is a one-to-onecorrespondence, given by the formulas (11.29) and (11.30), between numbers˛n 2 D and numbers snC1 in the interior of the disk Cs: This yields a one-to-onecorrespondence between numbers ˛n 2 D and extensions Qs D .s0; : : : ; snC1/ of sbelonging to the interior of the moment cone SnC2.

Remark 11.20

1. Set Os WD .s1; : : : ; snC1/. Then the determinant in (11.30) is just the determinantDn.Os/ for the sequence Os, that is, we have

˛n D .�1/nDn.Os/Dn.s/

:

Therefore, if � is a representing measure for s, then the (complex!) measure O�defined by d O�.z/ D z�1d�.z/ has the moments s1; : : : ; snC1:

2. We easily compute

˛0 D s1s0

and ˛1 D s0s2 � s21s20 � s1s�1

:

In fact, the reflection coeffients ˛n depend only on the quotients sjs0; j 2 N:

3. If s0 D 1, then (11.29) is a recursion formula which determines the momentsequence s uniquely in terms of the sequence .˛n/

1nD0. For this reason

we restricted ourselves to probability measures in Theorem 11.12 and inSect. 11.5. ı

11.5 Carathéodory Functions, the Schur Algorithm, and Geromimus’ Theorem 271

Proof of Theorem 11.12 (iii)!(i) Let ˛ D .˛n/n2N0 be a sequence of numbers˛n 2 D:By induction we construct a positive definite sequence s D .sj/j2N0 , s0 D 1,such that ˛n D ˛n.s/ for all n 2 N0.

Let n D 1 and set s WD .1; ˛0/. Then D1.s/ D 1 � j˛0j2 > 0, so that s 2 IntS2.Further, P0.z/ D P�

0 .z/ D 1 and P1.z/ D z � ˛0 D zP0 � ˛0P�0 :

Suppose now that sŒn� D .sj/njD0 2 IntSnC1 is constructed such that it has thereflection coefficients ˛0; : : : ; ˛n�1. Then, by Theorem 11.19, there exists an snC1 2C such that sŒnC1� D .s0; : : : ; snC1/ 2 IntSnC2 has the n-th reflection coefficient˛n. By induction the preceding gives the desired positive definite sequence s. ut

Summarizing the main results of this and the preceding sections we haveestablished a one-to-one correspondence between the following three objects:

� probability measures � on T of infinite support,� positive definite sequences s D .sj/j2Z on Z, where s0 D 1,� sequences ˛ D .˛n/n2N0 of complex numbers ˛n 2 D.

Indeed, for the probability measure � on T, s is its moment sequence (givenby sj D

R��jd�.�/, where j 2 Z). For the positive definite sequence s, ˛ is its

sequence of reflection coefficients from Theorem 11.9 (given by (11.30)) and � isthe unique solution of the moment problem for s from Theorem 11.3. Finally, forthe sequence ˛, s is the positive definite sequence from Theorem 11.12 (definedinductively by (11.29) and s0 WD 1).

11.5 Carathéodory Functions, the Schur Algorithm,and Geromimus’ Theorem

The following two notions on holomorphic functions are crucial in this section.

Definition 11.21 A holomorphic function f on D D fz 2 C W jzj < 1g is called a

� Carathéodory function if f .0/ D 1 and Re f .z/ � 0 for z 2 D,� Schur function if j f .z/j � 1 for z 2 D.

For � 2 T, the constant function f .z/ D � is obviously a Schur function. For allother Schur functions f we have j f .z/j < 1 on D by the maximum principle.

The next result is Herglotz’ representation theorem of Carathéodory functions.

Proposition 11.22 For each probability measure � on T, the function F�defined by

F�.z/ DZT

� C z

� � zd�.�/; z 2 D; (11.31)

is a Carathéodory function. Each Carathéodory function is of the form F� and theprobability measure � is uniquely determined by the function F�.


It is easily verified that F� is a Carathéodory function. Indeed, F� is holomorphicon D, F�.0/ D �.T/ D 1 and ReF�.z/ > 0 on D, since

Re� C z

� � zD 1 � jzj2j� � zj2 > 0; z 2 D; � 2 T:

That each Carathéodory function of the form (11.31) is proved (for instance) in [Dn,Theorem III on p. 21].

Using the representation (11.31) from Proposition 11.22 it is easy to relate theTaylor coefficients of Carathéodory functions to moment sequences.

Proposition 11.23 Let F be a holomorphic function on D with Taylor expansion

F.z/ D 1C 21XnD1

cnzn: (11.32)

Then F is a Carathéodory function if and only if there is a probability measure � onT such that

cn D sn.�/ �ZT

��nd�.�/ for n 2 N:

Proof Let F be a Carathéodory function. By Proposition 11.22, F is of the form(11.31). For z 2 D and � 2 T we have the expansion

� C z

� � zD 1C 2

1XnD1

��nzn (11.33)

which converges uniformly on T. Hence, since �.T/ D 1, integrating over T gives

F.z/ D 1C 21XnD0

sn.�/zn; z 2 D:

Comparing the coefficients of zn yields cn D sn.�/ for n 2 N.Conversely, suppose that cn D sn.�/, n 2 N, for some probability measure �.

Then, using (11.32) and the uniformly converging expansion (11.33) on T we obtain

F.z/ D 1C1XnD0

2

ZT

��nd�.�/ zn DZT

�1C 2

1XnD1

��nzn�d�.�/

DZT

� C z

� � zd�.�/;

that is, F D F�. Hence F is a Carathéodory function. ut


The following simple fact will be used several times.

Lemma 11.24 If f is a Schur function such that f .0/ D 0, then f .z/z is also a Schur

function.

Proof Since f .0/ D 0, Schwarz’ lemma applies and shows that j f .z/j � jzj forz 2 D. Hence j f .z/z j � 1 on D, so that f .z/

z is a Schur function. utNext we note that the map

f 7! F WD 1C zf .z/

1� zf .z/(11.34)

is a bijection of the Schur functions f onto the Carathéodory functions F with inverse

F 7! f D 1

z

F.z/� 1F.z/C 1 : (11.35)

Indeed, since w 7! v D 1Cw1�w is a holomorphic bijection of the open unit disc D on

the open half plane Re v > 0, the function F in (11.34) is a Carathéodory functionif f is a Schur function. The inverse v 7! w D v�1

vC1 maps the half plane Re v > 0

holomorphically onto D. Hence, if F is a Carathéodory function, the function zf .z/defined by (11.35) is a Schur function and so is f by Lemma 11.24.

If F� is the Carathéodory function of a probability measure �, we denote by

f� WD 1

z

F�.z/� 1F�.z/C 1 (11.36)

the corresponding Schur function. By the preceding we have developed one-to-onecorrespondences between probability measures on T, Carathéodory functions andSchur functions.

Recall that for any � 2 D the Möbius transformation

M� .w/ WD w � �1 � �z ; w 2 D;

is a holomorphic bijection of D and a bijection of T. Hence, finite Blaschkeproducts

f .z/ D ei'nY

jD1

z � �j1 � �jz

(11.37)

of order n, where ' 2 R and �1; : : : ; �n 2 D, are Schur functions. Constantfunctions of modulus one are interpreted as Blaschke products of order 0.

Lemma 11.25 For a probabilitiy measure � on T the following are equivalent:

(i) � has finite support.(ii) F� is a rational function with all its poles in T.

(iii) f� is a finite Blaschke product.


Proof Clearly, the points of supp� are the singularities of F�. This fact implies theequivalence of (i) and (ii).

From (11.34) and (11.35) it follows that f� is rational if and only if F� is rational;in this case f� has boundary values of modulus one on T. Hence (ii) holds if andonly if f� is an rational inner function. It is well-known (see e.g. [Y, p. 208]) thatthe rational inner functions are precisely the Blaschke products of finite order. ut

Let S denote the set of Schur functions and Sf the finite Blaschke products.Now we begin to develop the Schur algorithm. Let f be a fixed Schur function.

We define inductively f0.z/ WD f .z/ and

�n WD fn.0/; fnC1.z/ WD z�1M�n. fn.z// Dfn.z/� �n

z.1 � �nfn.z//; n 2 N0: (11.38)

Conversely, from (11.38) we obtain

fn.z/ D M��n.zfnC1/ D �n C zfnC1.z/1C �nzfnC1.z/

D �n C .1 � j�nj2/ zfnC1.z/1C �nzfnC1.z/

:

Suppose that �n 2 D and fn 2 S. Since M�n is a holomorphic bijection of D,M�n. fn/ 2 S and so fnC1 D z�1M�n. fn/ 2 S by Lemma 11.24. By induction thisproves that the functions fn.z/ are Schur functions if j�kj < 1 for k � n.

Assume in the above algorithm that �k 2 D for k D 0; : : : ; n � 1 and �n D fn.0/is not in D. Then �n 2 @D D T, the algorithm terminates, and j�kj D 1 for k � n. Itis easy to verify that this happens if and only if f 2 Sf .

Definition 11.26 The numbers �n D �n. f /, n 2 N0, are called the Schurparameters of the Schur function f .

Thus, if f 2 S and f … Sf , the Schur algorithm yields a sequence . fn/n2N0 ofSchur functions fn 2 SnSf and a sequence .�n/n2N0 of Schur parameters �n 2 D:

We shall write �n.�/ for the Schur parameters of the Schur function f� givenby (11.36). If the probability measure � has infinite support, then f� … Sf byLemma 11.25 and hence all Schur parameters �n.�/, n 2 N0, are in D.

For a Schur function f we denote by an. f / its n-th Taylor coefficient, that is,

f .z/ D1XnD0

an. f /zn:

It can be shown [Su] that the Schur parameter �n. f / is a function of the Taylorcoeffients a0. f /; : : : ; an. f / and that the Taylor coefficient ak. f / is a function of theSchur parameters �0. f /; : : : ; �k. f /. We shall use only the following results.

Proposition 11.27 For n 2 N there exists a real polynomial'n of 2n variables suchthat for any Schur function f we have

an. f / D �nn�1YjD0.1 � j�jj2/C 'n.�0; �0; : : : ; �n�1; �n�1/: (11.39)


Proof We prove by induction on n that for all k 2 N0 and n 2 N there exists apolynomial 'n;k in �j and � j for j D 0; : : : ; kC n � 1 such that

an. fk/ D �nCk

nCk�1YjDk

.1 � j�jj2/C 'n;k.�0; �0; : : : ; �nCk�1; �nCk�1/: (11.40)

Since f D f0, the case k D 0 gives the assertion.We compare the coefficient of z in the equation fk C � kzfkfkC1 D �k C zfkC1

by using that fj.0/ D �j. This yields a1. fk/ C � k�k�kC1 D �kC1. Hence we havea1. fk/ D �kC1.1 � j�kj2/. This is the assertion (11.40) for n D 1 and k 2 N0.

Assume that (11.40) holds for n � 1. Comparing the coefficient of zn in theidentity

fk.z/ D �k C zfkC1.z/� � kzfkC1.z/fk.z/

by using the definition a0. fk/ D fk.0/ D �k we derive

an. fk/ D an�1. fkC1/� � kan�1. fkC1fk/

D an�1. fkC1/� � kn�1XjD0

an�1�j. fkC1/aj. fk/

D .1 � j�kj2/an�1. fkC1/ � � kn�1XjD1

an�1�j. fkC1/aj. fk/: (11.41)

For the terms an�1. fkC1/, an�1�j. fkC1/, aj. fk/ in (11.41) the induction hypothesis(11.40) applies with n replaced by n�1. Hence an. fk/ is of the required form (11.40).This completes the induction proof. utCorollary 11.28 Suppose that f and g are Schur functions such that �j. f / D �j.g/for j D 0; : : : ; n. Then

j f .z/ � g.z/j � 2jzjnC1 for z 2 D: (11.42)

If f ; g 2 S have the same Schur parameters �k. f / D �k.g/ for k 2 N0, then f D g.

Proof First we note that f .0/ D �0. f / D �0.g/ D g.0/. By Proposition 11.27,for any j 2 N the j-th Taylor coefficient of a Schur function is a polynomial in itsSchur parameters �0; �0; : : : ; �j; � j. Hence the assumption implies that the first nC1Taylor coefficients of f and g coincide, so the Schur function h WD 1

2. f � g/ has a

zero of order nC1 at the origin. Therefore, by repeated application of Lemma 11.24we obtain jh.z/j � jzjnC1 on D which gives (11.42).

If �k. f / D �k.g/ for k 2 N0, then (11.42) holds for all n 2 N. Passing to thelimit n!1 yields f .z/ D g.z/ for z 2 D. ut


Now we reverse the procedure and start with Schur parameters. For � 2 D; set

T�;z.w/ WD � C zw

aC � zw :

Then T�;z. f .z// is also a Schur function for any f 2 S. From (11.38) we obtain

fn.z/ D T�n ;z. fnC1.z// D �n C zfnC1.z/1C �n zfnC1.z/

:

Let .�n/n2N0 be an arbitrary sequence of numbers �n 2 D. We develop the inverseSchur algorithm and define the n-th Schur approximant f Œn�.z/ by

f Œn�.z/ D T�0;z.T�1;z.: : : T�n�1;z.�n///; n 2 N0:

(In fact, this is a kind of continued fraction algorithm.) It is not difficult to verifythat f Œn�.z/ a rational Schur function with Schur parameters given by

�j. fŒn�/ D �j for j D 0; : : : ; n and �j. f

Œn�/ D 0 for j > n: (11.43)

Therefore, if n > m, it follows from by Corollary 11.28 that

j f Œn�.z/ � f Œm�.z/j � 2jzjnC1; z 2 D:

Hence . f Œn�.z//2N0 is a Cauchy sequence for fixed z 2 D which converges uniformlyon compact subsets of D to some holomorphic function f .z/ on D. Since f Œn� 2 S,it is obvious that f 2 S. Further, �k. f / D �k. f Œn�/ D �k for n � k, that is, the Schurfunction f has the prescribed Schur parameters �k; k 2 N0. Since �k 2 D for allk 2 N0, f is not in Sf . Therefore, by Lemma 11.25, the unique probability measure� such that f D f� has infinite support.

Now let g 2 SnSf be given. Then �n WD �n.g/ 2 D for n 2 N0. If we start theinverse Schur algorithm with this sequence .�n/n2N0 , then the corresponding Schurfunction f .z/ has the same Schur parameters as g.z/, so it coincides with g.z/ byCorollary 11.28. Hence the sequence . f Œn�.z//2N0 of Schur approximants convergesto the Schur function g.z/ uniformly on compact subsets of D.

For later reference we state an outcome of the preceding considerations as

Proposition 11.29 For each sequence .�n/n2N0 of numbers �n 2 D there exists aunique probability measure � on T with infinite support such that

�k.�/ � �k. f�/ D �k for k 2 N0:

The next proposition is needed in the proof of Theorem 11.31 below.

Proposition 11.30 Let � be a probability measure on T and let �j be its Schurparameters. For n 2 N0 there exists a real polynomial in 2n variables such that

snC1.�/ D �nn�1YjD0.1 � j�jj2/C n.�0; �0; : : : ; �n�1; �n�1/: (11.44)


Proof Recall that F� denotes the Carathéodory function and f� is the Schur functionassiciated with �. By (11.34) we then have

F�.z/ D 1C zf�.z/

1 � zf�.z/D 1C 2zf�.z/

1 � zf�.z/D 1C

1XnD1

2. f�z/n:

This formula implies that the Taylor coefficient anC1.F�/ is a sum of the number2an. f�/ and a polynomial in the lower coefficients aj. f�/, where j � n�1. On theother hand, anC1.F�/ D 2snC1.�/ by Proposition 11.23. Applying formula (11.39)to the Taylor coefficients an. f�/ and aj. f�/; j � n � 1, we obtain the assertion. ut

The main result of this section is the following Geronimus theorem.

Theorem 11.31 For each probability measure � on T of infinite support we have

˛n.�/ D �n.�/ for n 2 N0:

Proof The assertion will be proved by induction on n. For n D 0 it is easily checkedthat s1.�/ D ˛0.�/ D �0.�/.

Now suppose that ˛j.�/ D �j.�/ for j D 0; : : : ; n� 1. We fix these numbers andabbreviate them by ˛j. Recall that ˛j 2 D for each j.

By Verblunsky’s Theorem 11.12, for each number � 2 D there is a probabilitymeasure � with infinite support such that ˛n.�/ D � and ˛j.�/ D ˛j for j D0; : : : ; n � 1. By (11.29) the corresponding moment snC1.�/ is of the form

snC1.�/ D ˛n.�/n�1YjD0.1 � j˛jj2/C cnC1; (11.45)

where cnC1 depends only on the moments s0; : : : ; sn and so on ˛1; : : : ; ˛n�1 byformula (11.44). Note that cnC1 does not depend on ˛n.�/ D �.

On the other hand, by Proposition 11.29, for each � 2 D there exists a probabilitymeasure Q� such that �n. Q�/ D � and �j. Q�/ D ˛j for j D 0; : : : ; n � 1. ByProposition 11.30 the corresponding moment snC1. Q�/ is given by

snC1. Q�/ D �n. Q�/n�1YjD0.1 � j˛jj2/C n; (11.46)

where n depends only on ˛0; : : : ; ˛n�1, but not on �n. Q�/ D �.Formulas (11.45) and (11.46) describe the sets of possible .n C 1/-th moments

when ˛n.�/ D � and �n. Q�/ D �, respectively, run through D. Both sets are opendisks with centers cnC1 and n. Since these sets are the same, cnC1 D n:


Since ˛j.�/ D �j.�/ D ˛j for j D 0; : : : ; n � 1, we can set � D Q� D � inthe preceding formulas. Comparing the moment snC1.�/ in (11.45) and (11.46) byusing that cnC1 D n we obtain

˛n.�/

n�1YjD0.1 � j˛jj2/ D �n.�/

n�1YjD0.1 � j˛jj2/:

Since j˛jj < 1, this yields ˛n.�/ D �n.�/. This completes the induction proof. ut

11.6 Exercises

1. Let k 2 N and s D .sj/j2Z, where s0 D 1; sk�nDsn�k D 12; sj D 0 otherwise.

a. Show that s is a moment sequence for Z by “guessing” the representingmeasure.

b. Compute the Toeplitz determinants Hn.s/; n 2 N0.

2. Let g.�/ D PnlD0 al cos l� be a trigonometric “cosine polynomial” such that

g.�/ � 0 for � 2 Œ��; ��. Show that there exists a polynomial q.z/ DPnjD0 cjzj

with real coefficients cj such that g.�/ D jq.ei�/j2 for � 2 Œ��; ��.Hint: Show that if zj is a nonreal zero with multiplicity kj of the polynomial f .z/in the proof of Theorem 11.1, then so is its conjugate zj.

3. ([AK]) Let s D .sj/njD0, where s0 > 0 and n 2 N, be a real sequence. Define rk D12k

PkjD0

�kj

�sk�2j for k D 0; : : : ; n: Show that the following are equivalent:

(i)

nXj;kD0

sj�k ckcj � 0 for .c0; : : : ; cn/T 2 CnC1:

(ii) There are numbers tj 2 Œ0; ��; mj � 0 for j D 1; : : : ; l; l � 1C n2; such that

sk DlX

jD1mj cos ktj; k D 0; : : : ; n:

(iii) r D .rk/nkD0 is a truncated Œ�1; 1�-moment sequence.

Hint: rk D Ls...zC z�1/=2/k/:4. Formulate and prove the counterpart of Exercise 3 for an infinite real sequence

s D .sj/j2N0 .

More details and further results on Exercises 3 and 4 can be found in [AK, Theorems13–17].

11.7 Notes 279

In the following exercises we elaborate on the moment problem for a subarc of T,see [KN, p. 294–295]. Suppose that ��

2� ˛ < ˇ � �

2. Let T˛;ˇ denote the subset

of numbers ei t, where t 2 Œ2˛; 2ˇ�, of T. We abbreviate a D tan ˛, b D tanˇ,c D cos˛, d D cosˇ and define a Laurent polynomial '˛;ˇ by

'˛;ˇ.z/ D �2 cos.˛ � ˇ/C e.˛Cˇ/i z�1 C e�.˛Cˇ/i z:

5. Show that '˛;ˇ.z/ � 0 for z 2 T˛;ˇ and '˛;ˇ.z/ < 0 for z 2 TnT˛;ˇ :

Hint: Verify that '˛;ˇ.ei t/ D 4 sin.ˇ � t2/ sin. t

2� ˛/.

6. Show that if p 2 CŒz; z�1� is nonnegative on T˛;ˇ , then there exist polynomialsq1; q2 2 CŒz� such that p.z/ D jq1.z/j2 C '˛;ˇ.z/jq2.z/j2 for z 2 T:

Hints: First suppose��2< ˛ < ˇ < �

2. Set z D .1� ix/.1C ix/�1 and verify that

'˛;ˇ.z/ D 4.cd/�1.b � x/.x � a/.1 C x2/�1. Show that there exists a q 2 RŒx�and n 2 N such that p.z/ D q.x/.1C x2/n and q.x/ � 0 for x 2 Œa; b�.

Now let ��2< ˛ < ˇ D �

2: Set z D .1 C i.x � a//.1 � i.x � a//�1. Show

that '˛;ˇ.z/ D 4c�1x.1C .x� a/2/�1 and there are q 2 RŒx� and n 2 N such thatp.z/ D q.x/.1C .x � a/2/n and q.x/ � 0 for x 2 RC.

Apply Corollaries 3.24 and 3.25, respectively, to q.7. (Moment problem on a circular arc)

Let s D .sn/n2N0 be a complex sequence and define a sequence Os WD .Osn/n2N0 by

Osn WD e.˛Cˇ/i snC1 � 2 cos.˛ � ˇ/ sn C e�.˛Cˇ/i sn�1; n 2 Z:

(Recall that s�n WD sn for n < 0.) Verify that Ls.'˛;ˇf / D LOs. f / for f 2 CŒz; z�1�:Prove that s is a T˛;ˇ- moment sequence for Z, that is, there is a Radon

measure � on T supported on T˛;ˇ such that sn DRTz�n d� for n 2 Z; if and

only if the two infinite Toeplitz matrices H.s/ and H.Os/ are positive semidefinite.

11.7 Notes

The Fejér–Riesz Theorem 11.1 was proved in [Fj] and [Rz1]. The solution of thetrigonometric moment problem in the present form is due to O. Toeplitz [To].The Szegö recurrence relations (11.15) and (11.16) were obtained in Szegö [Sz],while Verblunsky’s Theorem 11.12 was proved in [Ver]. The circle Cs and the resultson the truncated trigonometric moment problem are due to N.I. Akhiezer and M.G.Krein [AK].

Carathéodory functions were first studied in [Ca1], [Ca2]. The Herglotz repre-sentation (Proposition 11.22) was obtained in [Hz]. Schur functions and Schur’salgorithm were invented in I. Schur’s two pioneering papers [Su], while Geronimus’Theorem 11.31 was proved in [Gs]. B. Simon’s book [Sim2] has several proofs ofVerblunsky’s and Geronimus’ theorems; the proof of Geronimus’ theorem given inthe text is taken from [Sim2]. Concerning Schur analysis, a collection of classicalpapers is [FK], a very readable discussion with many historical comments is [DK],and a deeper analysis is given in [Kv].

Part IIIThe Multidimensional Moment Problem

Chapter 12The Moment Problem on CompactSemi-Algebraic Sets

In this chapter we begin the study of the multidimensional moment problem. Thepassage to dimensions d � 2 brings new difficulties and unexpected phenomena. InSect. 3.2 we derived solvability criteria of the moment problem on intervals in termsof positivity conditions. It seems to be natural to look for similar characterizations inhigher dimensions as well. This leads us immediately into the realm of real algebraicgeometry and to descriptions of positive polynomials on semi-algebraic sets. In thischapter we treat this approach for basic closed compact semi-algebraic subsets ofRd. It turns out that for such sets there is a close interaction between the momentproblem and Positivstellensätze for strictly positive polynomials.

All basic notions and facts from real algebraic geometry that are needed for ourtreatment of the moment problem are collected in Sect. 12.1. Section 12.2 containsgeneral facts on localizing functionals and supports of representing measures.The main existence results for the moment problem (Theorems 12.25, 12.36(ii),and 12.45) and the corresponding Positivstellensätze (Theorems 12.24, 12.36(i),and 12.44) for compact semi-algebraic sets are derived in Sects. 12.3, 12.4, and 12.6.The results in Sects. 12.3 and 12.4 are formulated in the language of prerorderingsand quadratic modules, that is, in terms of weighted sums of squares. In Sect. 12.6we use another type of positivity condition which is based on the notion of asemiring.

In Sect. 12.4 we develop a fundamental technical result, the representationtheorem for Archimedean quadratic modules and semirings (Theorem 12.35). InSects. 12.6 and 12.7, the main theorems are applied to derive a number of classicalresults on the moment problem for concrete compact sets.

Apart from real algebraic geometry the theory of self-adjoint Hilbert spaceoperators is our main tool for the multidimensional moment problem. In Sect. 12.5we develop this method by studying the GNS construction and the relations to themultidimensional spectral theorem. This approach yields a very short and elegantproof of the moment problem result for Archimedean quadratic modules.


283

284 12 The Moment Problem on Compact Semi-Algebraic Sets

Throughout this chapter, A denotes a commutative real algebra with unitelement denoted by 1. For notational simplicity we write � for � � 1, where � 2 R.Recall that

PA2 is the set of finite sums

Pi a2i of squares of elements ai 2 A.

12.1 Semi-Algebraic Sets and Positivstellensätze

The following definition collects some basic notions needed in the sequel.

Definition 12.1 A quadratic module of A is a subset Q of A such that

QC Q Q; 1 2 Q; a2Q 2 Q for all a 2 A: (12.1)

A quadratic module T is called a preordering if T � T T.A semiring is a subset S of A satisfying

SC S S; S � S S; � 2 S for all � 2 R; � � 0: (12.2)

A cone is a subset C of A such that CC C C and � � C C for � � 0.

In the literature “semirings” are also called “preprimes”. The name “quadraticmodule” stems from the last condition in (12.1) which means that Q is invariantunder multiplication by squares. Setting a D p�, this implies that � � Q Q for� � 0. Hence quadratic modules are cones. While semirings and preorderings areclosed under multiplication, quadratic modules are not necessarily. Semirings donot contain all squares in general. Clearly, a quadratic module is a preordering ifand only if it is a semiring. In this book, we work mainly with quadratic modulesand preorderings. Semirings will occur only in Theorems 12.35, 12.44, and 12.45below.

Example 12.2 The subset S D fPnjD0 ajxj W aj � 0; n 2 Ng of RŒx� is a semiring,

but not a quadratic module. Clearly, Q DPRdŒ x �2Cx1P

RdŒ x �2Cx2P

RdŒ x �2

is a quadratic module of RdŒ x �; d � 2, but Q is neither a semiring nor a preordering.ı

Each cone C of A yields an ordering� on A by defining

a � b if and only if b � a 2 C:

Obviously,P

A2 is the smallest quadratic module of A. Since A is commutative,PA2 is invariant under multiplication, so it is also the smallest preordering of A.Our guiding example for A is the polynomial algebra RdŒ x � WD RŒx1; : : : ; xd�.Let f D f f1; : : : ; fkg be a finite subset of RdŒ x �. The set

K.f/ � K. f1; : : : ; fk/ D fx 2 Rd W f1.x/ � 0; : : : ; fk.x/ � 0g (12.3)

12.1 Semi-Algebraic Sets and Positivstellensätze 285

is called the basic closed semi-algebraic set associated with f. It is easily seen that

Q.f/ � Q. f1; : : : ; fk/ D˚�0 C f1�1 C � � � C fk�k W �0; : : : ; �k 2

XRdŒ x �

2�

(12.4)

is the quadratic module generated by the set f and that

T.f/ � T. f1; : : : ; fk/ D� X

eD.e1;:::;ek/2f0;1gkf e11 � � � f ekk �e W �e 2

XRdŒ x �

2

�

(12.5)

is the preordering generated by the set f. These three sets K.f/, Q.f/, and T.f/ play acrucial role in this chapter and the next.

By the above definitions, all polynomials from T.f/ are nonnegative on K.f/, butin general T.f/ does not exhaust the nonnegative polynomials on K.f/.

The following Positivstellensatz of Krivine–Stengle is a fundamental result ofreal algebraic geometry. It describes nonnegative resp. positive polynomials on K.f/in terms of quotients of elements of the preordering T.f/.

Theorem 12.3 Let K.f/ and T.f/ be as above and let g 2 RdŒ x �. Then we have:

(i) (Positivstellensatz) g.x/ > 0 for all x 2 K.f/ if and only if there existpolynomials p; q 2 T.f/ such that pg D 1C q.

(ii) (Nichtnegativstellensatz) g.x/ � 0 for all x 2 K.f/ if and only if there existp; q 2 T.f/ and m 2 N such that pg D g2m C q.

(iii) (Nullstellensatz) g.x/ D 0 for x 2 K.f/ if and only if �g2n 2 T.f/ for somen 2 N.

(iv) K.f/ is empty if and only if �1 belongs to T.f/:Proof See [PD] or [Ms1]. The orginal papers are [Kv1] and [Ste1]. ut

All “if” assertions are easily checked and it is not difficult to show that all fourstatements are equivalent, see e.g. [Ms1]. Standard proofs of Theorem 12.3 as givenin [PD] or [Ms1] are based on the Tarski–Seidenberg transfer principle. Assertion (i)of Theorem 12.3 will play an essential role in the proof of Proposition 12.22 below.

Now we turn to algebraic sets. For a subset S of RdŒ x �; the real zero set of S is

Z.S/ D fx 2 Rd W f .x/ D 0 for all f 2 Sg: (12.6)

A subset V of Rd of the form Z.S/ is called a real algebraic set.Hilbert’s basis theorem [CLO, p. 75] implies that each real algebraic set is of the

form Z.S/ for some finite set S D fh1; : : : ; hmg. In particular, each real algebraic setis a basic closed semi-algebraic set, because K.h1; : : : ; hm;�h1; : : : ;�hm/ D Z.S/.

Let S be a subset of RdŒ x � and V WD Z.S/ the corresponding real algebraic set.We denote by I the ideal of RdŒ x � generated by S and by OI the ideal of f 2 RdŒ x �


which vanish on V . Clearly, Z.S/ D Z.I/ and I OI . In general, I ¤ OI. (Forinstance, if d D 2 and S D fx21 C x22g, then V D f0g and x21 2 OI , but x21 … I.)

It can be shown [BCRo, Theorem 4.1.4] that I D OI if and only ifP

p2j 2 Ifor finitely many pj 2 RdŒ x � implies that pj 2 I for all j. An ideal that obeys thisproperty is called real. In particular, OI is real. The ideal I generated by a singleirreducible polynomial h 2 RdŒ x � is real if and only if h changes its sign on Rd,that is, there are x0; x1 2 Rd such that h.x0/h.x1/ < 0, see [BCRo, Theorem 4.5.1].

The quotient algebra

RŒV� WD RdŒ x �= OI (12.7)

is called the algebra of regular functions on V . Since OI is real, it follows that

XRŒV�2 \ � �XRŒV�2

� D f0g: (12.8)

Example 12.4 Let us assume that the set f is of the form

f D fg1; � � � ; gl; h1;�h1; : : : ; hm;�hmg:

If g WD fg1; : : : ; glg and I denotes the ideal of RdŒ x � generated by h1; : : : ; hm, then

K.f/ D K.g/ \Z.I/; Q.f/ D Q.g/C I; and T.f/ D T.g/C I: (12.9)

We prove (12.9). The first equality of (12.9) and the inclusions Q.f/ Q.g/C Iand T.f/ T.g/C I are clear from the corresponding definitions. The identity

phj D 1

4Œ. pC 1/2hj C . p � 1/2.�hj/� 2 Q.f/; p 2 RdŒ x �;

implies that I Q.f/ T.f/. Hence Q.g/C I Q.f/ and T.g/C I T.f/: ıAnother important concept is introduced in the following definition.

Definition 12.5 Let Q be a quadratic module or a semiring of A. Define

Ab.Q/ WD fa 2 A W there exists a � > 0 such that � � a 2 Qand �C a 2 Qg:

We shall say that Q is Archimedean if Ab.Q/ D A, or equivalently, for every a 2 Athere exists a � > 0 such that � � a 2 A:

Lemma 12.6 Let Q be a quadratic module of A and let a 2 A. Then a 2 Ab.Q/ ifand only if �2 � a2 2 Q for some � > 0.

Proof If �˙ a 2 Q for � > 0, then

�2 � a2 D 1

2�

�.�C a/2.� � a/C .� � a/2.�C a/

� 2 Q:

12.1 Semi-Algebraic Sets and Positivstellensätze 287

Conversely, if �2 � a2 2 Q and � > 0, then

�˙ a D 1

2�

�.�2 � a2/C .�˙ a/2

� 2 Q: ut

Lemma 12.7 Suppose that Q is a quadratic module or a semiring of A.

(i) Ab.Q/ is a unital subalgebra of A.(ii) If the algebra A is generated by elements a1; : : : ; an, then Q is Archimedean if

and only if each ai there exists a �i > 0 such that �i ˙ ai 2 Q.

Proof

(i) Clearly, sums and scalar multiples of elements of Ab.Q/ are again in Ab.Q/. Itsuffices to verify that this holds for the product of elements a; b 2 Ab.Q/.

First we suppose that Q is a quadratic module. By Lemma 12.6, there are�1 > 0 and �2 > 0 such that �21 � a2 and �22 � b2 are in Q. Then

.�1�2/2 � .ab/2 D �22.�21 � a2/C a2.�22 � b2/ 2 Q;

so that ab 2 Ab.Q/ again by Lemma 12.6.Now let Q be a semiring. If �1 � a 2 Q and �2 � b 2 Q, then

�1�2 � ab D 1

2

�.�1 ˙ a/.�2 � b/C .�2 � a/.�2 C b/

� 2 Q:

(ii) follows at once from (i). utBy Lemma 12.7(ii), it suffices to check the Archimedean condition � ˙ a 2 Q

for algebra generators. Often this simplifies proving that Q is Archimedean.

Corollary 12.8 For a quadratic module Q of RdŒ x � the following are equiva-lent:

(i) Q is Archimedean.(ii) There exists a number � > 0 such that � �Pd

kD1 x2k 2 Q.(iii) For any k D 1; : : : ; d there exists a �k > 0 such that �k � x2k 2 Q.

Proof (i)!(ii) is clear by definition. If � �PdjD1 x2j 2 Q, then

� � x2k D � �X

jx2j C

Xj¤k

x2j 2 Q:

This proves (ii)!(iii). Finally, if (iii) holds, then xk 2 Ab.Q/ by Lemma 12.6 andhence Ab.Q/ D A by Lemma 12.7(ii). Thus, (iii)!(i). utCorollary 12.9 If the quadratic module Q.f/ ofRdŒ x � is Archimedean, then the setK.f/ is compact.


Proof By the respective definitions, polynomials of Q.f/ are nonnegative on K.f/.Since Q.f/ is Archimedean, � �Pd

kD1 x2k 2 Q.f/ for some � > 0 by Corollary 12.8,so K.f/ is contained in the ball centered at the origin with radius

p�. ut

The converse of Corollary 12.9 does not hold, as the following example shows.(However, it does hold for the preordering T.f/ as shown by Proposition 12.22below.)

Example 12.10 Let f1 D 2x1 � 1, f2 D 2x2 � 1, f3 D 1 � x1x2. Then the set K.f/ iscompact, but Q.f/ is not Archimedean (see [PD], p. 146, for a proof). ı

The following separation result is used several times in the next sections.

Proposition 12.11 Let Q be an Archimedean quadratic module of A. If a0 2 A anda0 … Q, there exists a Q-positive linear functional ' on A such that '.1/ D 1 and'.a0/ � 0.Proof Let a 2 A and choose � > 0 such that � ˙ a 2 Q. If 0 < ı � ��1, thenı�1 ˙ a 2 Q and hence 1 ˙ ıa 2 Q. This shows that 1 is an internal point of Q.Therefore, Eidelheit’s separation theorem (Theorem A.27) applies and there exists aQ-positive linear functional ¤ 0 on A such that .a0/ � 0. Since ¤ 0, we have .1/ > 0. (Indeed, if .1/ D 0, since is Q-positive, �˙a 2 Q implies .a/ D 0for all a 2 A and so D 0.) Then ' WD .1/�1 has the desired properties. utExample 12.12 Let A D RdŒ x � and let K be a closed subset of Rd. If Q is thepreordering Pos.K/ of nonnegative polynomials on K, then Ab.Q/ is just the set ofbounded polynomials on K. Hence Q is Archimedean if and only if K is compact. ı

Recall that OA denotes the set of characters of A. For a subset Q of A we define

K.Q/ WD fx 2 OA W f .x/ � 0 for all f 2 Qg: (12.10)

Clearly, if Q is the quadratic module Q.f/ of A D RdŒ x � defined by (12.4), thenK.Q/ is just the semi-algebraic set K.f/ given by (12.3).

Let Q be a quadratic module. The set Qsat D Pos.K.Q// of all f 2 A which arenonnegative on the set K.Q/ is obviously a preordering of A that contains Q. ThenQ is called saturated if Q D Qsat, that is, if Q is equal to its saturation Qsat.

Real algebraic geometry is treated in the books [BCRo, PD, Ms1]; a recent surveyon positivity and sums of squares is given in [Sr3].

12.2 Localizing Functionals and Supportsof Representing Measures

Haviland’s Theorem 1.12 shows that there is a close link between positive polyno-mials and the moment problem. However, in order to apply this result reasonabledesriptions of positive, or at least of strictly positive, polynomials are needed.

12.2 Localizing Functionals and Supports of Representing Measures 289

Recall that the moment problem for a functional L on the interval Œa; b� is solvableif and only if L. p2C .x�a/.b� x/q2/ � 0 for all p; q 2 RŒx�. This condition meansthat two infinite Hankel matrices are positive semidefinite and this holds if and onlyif all principal minors of these matrices are nonnegative. In the multidimensionalcase we are trying to find similar solvability criteria. It is natural to consider setsthat are defined by finitely many polynomial inequalities f1.x/ � 0; : : : ; fk.x/ � 0.These are precisely the basic closed semi-algebraic sets K.f/, so we have enteredthe setup of real algebraic geometry.

Let us fix a semi-algebraic set K.f/. Let L be a K.f/-moment functional, that is,L is of the form L. p/ D L�. p/ � R

p d� for p 2 RdŒ x �, where � is a Radonmeasure supported on K.f/. If g 2 RdŒx� is nonnegative on K.f/, then obviously

L.gp2/ � 0 for all p 2 RdŒ x �; (12.11)

so (12.11) is a neccesary condition for L being a K.f/-moment functional.The overall strategy in this chapter and the next is to solve the K.f/-moment

problem by finitely many sufficient conditions of the form (12.11). That is, our aimis to “find” nonnegative polynomials g1; : : : ; gm on K.f/ such that the followingholds:

Each linear functional L on RdŒ x � which satisfies condition (12.11) for g Dg1; : : : ; gm and g D 1 is a K.f/-moment functional. (The polynomial g D 1 isneeded in order to ensure that L itself is a positive functional.)

In general it is not sufficient to take only the polynomials fj themselves as gj. Forour main results (Theorems 12.25 and 13.10), the positivity of the functional on thepreordering T.f/ is assumed. This means that condition (12.11) is required for allmixed products g D f e11 � � � f ekk , where ej 2 f0; 1g for j D 1; : : : ; k.

Definition 12.13 Let L be a linear functional on RdŒ x � and let g 2 RdŒ x �. Thelinear functional Lg on RdŒ x � defined by Lg. p/ D L.gp/; p 2 RdŒ x �, is called thelocalization of L at g or simply the localized functional.

Condition (12.11) means the localized functionalLg is a positive linear functionalon RdŒ x �: Further, if L comes from a measure � supported on K.f/ and g isnonnegative on K.f/, then

Lg. p/ D L.gp/ DZK.f/

p.x/ g.x/d�.x/; p 2 RdŒ x �;

that is, Lg is given by the measure � on K.f/ defined by d� D g.x/d�.Localized functionals will play an important role throughout our treatment. They

are used to localize the support of the measure (see Propositions 12.18 and 12.19and Theorem 14.25) or to derive determinacy criteria (see Theorem 14.12).

Now we introduce two other objects associated with the functional L and thepolynomial g. Let s D .s˛/˛2Nd

0be the d-sequence given by s˛ D L.x˛/ and write


g DP� g�x� . Then we define a d-sequence g.E/s D ..g.E/s/˛/˛2Nd

0by

.g.E/s/˛ WDX

�g�s˛C� ; ˛ 2 Nd

0;

and an infinite matrix H.gs/ D .H.gs/˛;ˇ/˛;ˇ2Nd0

over Nd0 �Nd

0 with entries

H.gs/˛;ˇ WDX

�g� s˛CˇC� ; ˛; ˇ 2 Nd

0: (12.12)

Using these definitions for p.x/ DP˛ a˛x˛ 2 RdŒ x � we compute

Ls.gp2/ D

X˛;ˇ;�

a˛aˇg� s˛CˇC� DX˛;ˇ

a˛aˇ.g.E/s/˛Cˇ DX˛;ˇ

a˛aˇH.gs/˛;ˇ:

(12.13)

This shows that g.E/s is the d-sequence for the functional Lg and H.gs/ is a Hankelmatrix for the sequence g.E/s. The matrix H.gs/ is called the localized Hankelmatrix of s at g.

Proposition 12.14 Let Q.g/ be the quadratic module generated by the finite subsetg D fg1; : : : ; gmg of RdŒ x �. Let L be a linear functional onRdŒ x � and s D .s˛/˛2Nd

0

the d-sequence defined by s˛ D L.x˛/: Then the following are equivalent:

(i) L is a Q.g/-positive linear functional on RdŒ x �.(ii) L;Lg1 ; : : : ;Lgm are positive linear functionals on RdŒ x �.

(iii) s; g1.E/s; : : : ; gm.E/s are positive semidefinite d-sequences.(iv) H.s/;H.g1s/; : : : ;H.gms/ are positive semidefinite matrices.

Proof The equivalence of (i) and (ii) is immediate from the definition (12.4) of thequadratic module Q.g/ and Definition 12.13 of the localized functionals Lgj .

By Proposition 2.7, a linear functional is positive if and only if the correspondingsequence is positive semidefinite, or equivalently, the Hankel matrix is positivesemidefinite. By (12.13) this gives the equivalence of (ii), (iii), and (iv). ut

The solvabability conditions in the existence theorems for the moment problemin this chapter and the next are given in the form (i) for some finitely generatedquadratic module or preordering. This means that condition (12.11) is satisfied forfinitely many polynomials g. Proposition 12.14 says there are various equivalentformulations of these solvability criteria: They can be expressed in the language ofreal algebraic geometry (in terms of quadratic modules, semirings or preorderings),of �-algebras (as positive functionals on RdŒ x �), of matrices (by the positivesemidefiniteness of Hankel matrices) or of sequences (by the positive semidefinite-ness of sequences).

The next proposition contains a useful criterion for localizing supports ofrepresenting measures. We denote by MC.Rd/ the set of Radon measure � onRd for which all moments are finite, or equivalently,

R jp.x/j d� < 1 for allp 2 RdŒ x �.

12.2 Localizing Functionals and Supports of Representing Measures 291

Proposition 12.15 Let � 2 MC.Rd/ and let s be the moment sequence of �.Further, let gj 2 RdŒ x � and cj � 0 be given for j D 1; : : : ; k: Set

K D fx 2 Rd W jgj.x/j � cj for j D 1; : : : ; kg: (12.14)

Then we have supp � K if and only if there exist constants Mj > 0 such that

Ls.g2nj / � Mjc

2nj for n 2 N; j D 1; : : : ; k: (12.15)

Proof The only if part is obvious. We prove the if direction and slightly modify theargument used in the proof of Proposition 4.1.

Let t0 2 RdnK. Then there is an index j D 1; : : : ; k such that jgj.t0/j > cj. Hencethere exist a number � > cj and a ball U around t0 such that jgj.t/j � � for t 2 U.For n 2 N we then derive

�2n�.U/ �ZUgj.t/

2n d�.t/ �ZRd

gj.t/2n d�.t/ D Ls.g

2nj / � Mjc

2nj :

Since � > cj, this is only possible for all n 2 N if �.U/ D 0. Therefore, t0 …supp �. This proves that supp � K: ut

We state the special case gj.x/ D xj of Proposition 12.15 separately as

Corollary 12.16 Suppose that c1 > 0; : : : ; cd > 0. Let � 2 MC.Rd/ withmoment sequence s. Then the measure � is supported on the d-dimensionalinterval Œ�c1; c1� � � � � � Œ�cd; cd� if and only if there are positive constants Mj

such that

Ls.x2nj / � s2n.0;:::;0;1;0;:::;0/ � Mjc

2nj for n 2 N; j D 1; : : : ; d:

The following two propositions are basic results about the moment problemon compact sets. Both follow from Weierstrass’ theorem on approximation ofcontinuous functions by polynomials.

Proposition 12.17 If � 2 MC.Rd/ is supported on a compact set, then � isdeterminate. In particular, if K is a compact subset of Rd, then each K-momentsequence, so each measure � 2M.Rd/ supported on K, is determinate.

Proof Let � 2MC.Rd/ be a measure having the same moments and so the samemoment functional L as �. Fix h 2 Cc.R

d;R/. We choose a compact d-dimensionalinterval K containing the supports of � and h. From Corollary 12.16 it follows thatsupp � K. By Weierstrass’ theorem, there is a sequence . pn/n2N of polynomialspn 2 RdŒ x � converging to h uniformly on K. Passing to the limits in the equality

ZKpn d� D L. pn/ D

ZKpn d�

we getRh d� D R h d�. Since this holds for all h 2 Cc.R

d;R/, we have� D �. ut


Proposition 12.18 Suppose that � 2MC.Rd/ is supported on a compact set. Letf D f f1; : : : ; fkg be a finite subset of RdŒ x � and assume that the moment functionaldefined by L�. p/ D R p d�, p 2 RdŒ x �, is Q.f/-positive. Then supp � K.f/:

Proof Suppose that t0 2 RdnK.f/. Then there exist a number j 2 f1; : : : ; kg, a ball Uwith radius > 0 around t0, and a number ı > 0 such that fj � �ı on 2U. We definea continuous function h on Rd by h.t/ D p

2�jjt � t0jj for jjt � t0jj � 2 andh.t/ D 0 otherwise and take a compact d-dimensional interval K containing 2U andsupp �. By Weierstrass’ theorem, there is a sequence of polynomials pn 2 RdŒ x �converging to h uniformly on K. Then fjp2n ! fjh2 uniformly on K and hence

limn

L�.fjp2n/ D

ZK.lim

nfjp

2n/ d� D

ZKfjh

2 d� DZ2U

fj.t/.2�jjt � t0jj/ d�.t/

�Z2U�ı.2�jjt � t0jj/ d� � �

ZUı d�.t/ D �ı�.U/: (12.16)

Since L� is Q.f/-positive, we have L�. fjp2n/ � 0. Therefore, �.U/ D 0 by (12.16),so that t0 … supp �. This proves that supp � K.f/: ut

The assertions of Propositions 12.17 and 12.18 are no longer valid if thecompactness assumptions are omitted. But the counterpart of Proposition 12.18 forzero sets of ideals holds without any compactness assumption.

Proposition 12.19 Let � 2MC.Rd/ and let I be an ideal ofRdŒ x �. If the momentfunctional L� of � is I-positive, then L� annihilates I and supp� Z.I/.

(As usual, Z.I/ D fx 2 Rd W p.x/ D 0 for p 2 Ig is the zero set of I.)

Proof If p 2 I, then �p 2 I and hence L�.˙p/ � 0 by the I-positivity of L�, sothat L�. p/ D 0. That is, L� annihilates I.

Let p 2 I. Since p2 2 I, we have L�. p2/ D Rp2 d� D 0. There-

fore, from Proposition 1.23 it follows that supp� Z. p2/ D Z. p/. Thus,supp� Z.I/. ut

For a linear functional L on RdŒ x � we define

NC.L/ WD ff 2 Pos.Rd/ W L. p/ D 0 g:

Proposition 12.20 Let L be a moment functional on RdŒ x �, that is, L D L� forsome � 2 MC.Rd/. Then the ideal IC.L/ of RdŒ x � generated by NC.L/ isannihilated by L and the support of each representing measure of L is containedin Z.IC.L//.

Proof Let � be an arbitrary representing measure of L. If f 2 NC.L/, then wehave L. f / D R

f .x/ d� D 0. Since f 2 Pos.Rd/, Proposition 1.23 applies andyields supp � Z. f /. Hence supp � Z.NC.L/// D Z.IC.L//: In particular, theinclusion supp � Z.IC.L// implies that L D L� annihilates IC.L/: ut

12.3 The Moment Problem on Compact Semi-Algebraic Sets and the Strict. . . 293

12.3 The Moment Problem on Compact Semi-Algebraic Setsand the Strict Positivstellensatz

The solutions of one-dimensional moment problems have been derived from desrip-tions of nonnegative polynomials as weighted sums of squares. The counterpartsof the latter in the multidimensional case are the so-called “Positivstellensätze”of real algebraic geometry. In general these results require denominators (seeTheorem 12.3), so they do not yield reasonable criteria for solving momentproblems. However, for strictly positive polynomials on compact semi-algebraicsets K.f/ there are denominator free Positivstellensätze (Theorems 12.24 and 12.36)which provides solutions of moment problems. Even more, it turns out that there isa close interplay between this type of Positivstellensätze and moment problems oncompact semi-algebraic sets, that is, existence results for the moment problem canbe derived from Positivstellensätze and vice versa.

We state the main technical steps of the proofs separately as Propositions 12.21–12.23. Proposition 12.23 is also used in a crucial manner in the proof of Theo-rem 13.10 below.

Suppose that f D f f1; : : : ; fkg is a finite subset of RdŒ x �. Let B.K.f// denote thealgebra of all polynomials of RdŒ x � which are bounded on the set K.f/.

Proposition 12.21 Let g 2 B.K.f// and � > 0. If �2 > g.x/2 for all x 2 K.f/, thenthere exists a p 2 T.f/ such that

g2n � �2nC2p for n 2 N: (12.17)

Proof By the Krivine–Stengle Positivstellensatz (Theorem 12.3(i)), applied to thepositive polynomial �2 � g2 on K.f/, there exist polynomials p; q 2 T.f/ such that

p.�2 � g2/ D 1C q: (12.18)

Since q 2 T.f/ and T.f/ is a quadratic module, g2n.1 C q/ 2 T.f/ for n 2 N0.Therefore, using (12.18) we conclude that

g2nC2p D g2n�2p � g2n.1C q/ � g2n�2p:

By induction it follows that

g2np � �2np: (12.19)

Since g2n.qC pg2/ 2 T.f/, using first (12.18) and then (12.19) we derive

g2n � g2n C g2n.qC pg2/ D g2n.1C qC pg2/ D g2n�2p � �2nC2p : ut


Proposition 12.22 If the set K.f/ is compact, then the associated preordering T.f/is Archimedean.

Proof Put g.x/ WD .1Cx21/ � � � .1Cx2d/. Since g is bounded on the compact set K.f/,we have �2 > g.x/2 on K.f/ for some � > 0. Therefore, by Proposition 12.21 thereexists a p 2 T.f/ such that (12.17) holds.

Further, for any multiindex ˛ 2 Nd0, j˛j � k, k 2 N, we obtain

˙2x˛ � x2˛ C 1 �Xjˇj�k

x2ˇ D gk: (12.20)

Hence there exist numbers c > 0 and k 2 N such that p � 2cgk. Combining thelatter with g2n � �2nC2p by (12.17), we get g2k � �2kC22cgk and so

.gk��2kC2c/2 � .�2kC2c/2�1:

Hence, by Lemma 12.6, gk��2kC2c 2 Ab.T.f// and so gk 2 Ab.T.f//, where A WDRdŒ x �. Since ˙xj � gk by (12.20) and gk 2 Ab.T.f//, we obtain xj 2 Ab.T.f// forj D 1; � � �; d. Now from Lemma 12.7(ii) it follows that Ab.T.f// D A. This meansthat T.f/ is Archimedean. utProposition 12.23 Suppose that L is a T.f/-positive linear functional onRdŒ x �.

(i) If g 2 B.K.f// and kgk1 denotes the supremum of g on K.f/; then

jL.g/j � L.1/ kgk1: (12.21)

(ii) If g 2 B.K.f// and g.x/ � 0 for x 2 K.f/, then L.g/ � 0.Proof

(i) Fix " > 0 and put � WDk g k1 C". We define a real sequence s D .sn/n2N0 bysn WD L.gn/. Then Ls.q.y// D L.q.g// for q 2 RŒy�. For any p 2 RŒy�, we havep.g/2 2 PRdŒ x �2 T.f/ and hence Ls. p.y/2/ D L. p.g/2/ � 0, since L isT.f/-positive. Thus, by Hamburger’s theorem 3.8, there exists a Radon measure� on R such that sn D

RRtnd�.t/, n 2 N0.

For � > � let �� denote the characteristic function of the set .�1;�� [Œ�;C1/. Since �2 � g.x/2 > 0 on K.f/, we have g2n � �2nC2p by Eq. (12.17)in Proposition 12.21. Using the T.f/-positivity of L we derive

�2nZR

�� .t/ d�.t/ �ZR

t2nd�.t/ D s2n D L.g2n/ � �2nC2L. p/ (12.22)

for all n 2 N. Since � > �, (12.22) implies thatRR��.t/ d�.t/ D 0. Therefore,

supp � Œ��; ��. (The preceding argument has been already used in the proofof Proposition 12.15 to obtain a similar conclusion.) Therefore, applying the


Cauchy–Schwarz inequality for L we derive

jL.g/j2 � L.1/L.g2/ D L.1/s2 D L.1/Z �

��t2 d�.t/

� L.1/�.R/�2 D L.1/2�2 D L.1/2.kgk1 C "/2:Letting "! C0, we get jL.g/j � L.1/ k g k1.

(ii) Since g � 0 on K.f/, we clearly have k 1 � kgk1 � 2 gk1 D kgk1: Using thisequality and (12.21) we conclude that

L.1/kgk1 � 2 L.g/ D L.1 � kgk1 � 2 g/ � L.1/k1 � kgk1 � 2 gk1 D L.1/kgk1;

which in turn implies that L.g/ � 0. utThe following theorem is the strict Positivstellensatz for compact basic closed

semi-algebraic sets K.f/:

Theorem 12.24 Let f D f f1; : : : ; fkg be a finite subset ofRdŒ x � and let h 2 RŒx�. Ifthe set K.f/ is compact and h.x/ > 0 for all x 2 K.f/, then h 2 T.f/.

Proof Assume to the contrary that h is not in T.f/. By Proposition 12.22, T.f/ isArchimedean. Therefore, by Proposition 12.11, there exists a T.f/-positive linearfunctional L on A such that L.1/ D 1 and L.h/ � 0. Since h > 0 on the compactset K.f/, there is a positive number ı such that h.x/ � ı > 0 for all x 2 K.f/. Weextend the continuous function

ph.x/� ı on K.f/ to a continuous function on some

compact d-dimensional interval containing K.f/. Again by the classical Weierstrasstheorem,

ph.x/� ı is the uniform limit on K.f/ of a sequence . pn/ of polynomials

pn 2 RdŒ x �. Then p2n�hCı ! 0 uniformly onK.f/, that is, limn k p2n�hCı k1D 0.Recall that B.K.f// D RdŒ x �, since K.f/ is compact. Hence limn L. p2n�hC ı/ D 0by the inequality (12.21) in Proposition 12.23(i). But, since L. p2n/ � 0, L.h/ � 0,and L.1/ D 1, we have L. p2n � hC ı/ � ı > 0 which is the desired contradiction.This completes the proof of the theorem. ut

The next result gives a solution of the K.f/-moment problem for compact basicclosed semi-algebraic sets.

Theorem 12.25 Let f D f f1; : : : ; fkg be a finite subset of RdŒ x �. If the set K.f/ iscompact, then each T.f/-positive linear functional L on RdŒ x � is a K.f/-momentfunctional.

Proof Since K.f/ is compact, B.K.f// D RdŒ x �. Therefore, it suffices to combineProposition 12.23(ii) with Haviland’s Theorem 1.12. utRemark 12.26 Theorem 12.25 was obtained from Proposition 12.23(ii) and Hav-iland’s Theorem 1.12. Alternatively, it can derived from Proposition 12.23(i)combined with Riesz’ representation theorem. Let us sketch this proof. By (12.21),the functional L on RdŒ x � is k � k1- continuous. Extending L to C.K.f// by theHahn–Banach theorem and applying Riesz’ representation theorem for continuouslinear functionals, L is given by a signed Radon measure on K.f/. Setting g D 1 in


(12.21), it follows that L, hence the extended functional, has the norm L.1/. It is notdifficult to show that this implies that the representing measure is positive. ı

The shortest path to Theorems 12.24 and 12.25 is probably to use Proposi-tion 12.23 as we have done. However, in order to emphasize the interaction betweenboth theorems and so in fact between the moment problem and real algebraicgeometry we now derive each of these theorems from the other.

Proof of Theorem 12.25 (Assuming Theorem 12.24) Let h 2 RdŒ x �. If h.x/ > 0

on K.f/, then h 2 T.f/ by Theorem 12.24 and so L.h/ � 0 by the assumption.Therefore L is a K.f/-moment functional by the implication (ii)!(iv) of Haviland’sTheorem 1.12. utProof of Theorem 12.24 (Assuming Theorem 12.25 and Proposition 12.22) Sup-pose h 2 RdŒ x � and h.x/ > 0 on K.f/. Assume to the contrary that h … T.f/.Since the preordering T.f/ is Archimedean by Proposition 12.22, Proposition 12.11applies, so there is a T.f/-positive linear functional L on RdŒ x � such that L.1/ D 1

and L.h/ � 0. By Theorem 12.25, L is a K.f/-moment functional, that is, thereis a measure � 2 MC.K.f// such that L. p/ D R

K.f/ p d� for p 2 RdŒ x �. ButL.1/ D �.K.f// D 1 and h > 0 on K.f/ imply that L.h/ > 0. This is a contradiction,since L.h/ � 0. ut

The preordering T.f/ was defined as the sum of sets f e11 � � � f ekk �P

RdŒ x �2. It isnatural to ask whether or not all such sets with mixed products f e11 � � � f ekk are reallyneeded. To formulate the corresponding result we put lk WD 2k�1 and let g1; : : : ; glkdenote the first lk polynomials of the following row of mixed products:

f1; : : : ; fk; f1 f2; f1 f3; : : : ; f1 fk; : : : ; fk�1 fk; f1 f2 f3; : : : ; fk�2 fk�1 fk; : : : ; f1 f2 : : : ; fk:

Let Q.g/ denote the quadratic module generated by g1; : : : ; glk , that is,

Q.g/ WDX

RdŒ x �2 C g1

XRdŒ x �

2 C � � � C glkX

RdŒ x �2:

The following result of T. Jacobi and A. Prestel [JP] sharpens Theorem 12.24.

Theorem 12.27 If the set K.f/ is compact and h 2 RdŒ x � satisfies h.x/ > 0 for allx 2 K.f/, then h 2 Q.g/.

We do not prove Theorem 12.27; for a proof of this result we refer to [JP]. If wetake Theorem 12.27 for granted and combine it with Haviland’s theorem 1.12 weobtain the following corollary.

Corollary 12.28 If the set K.f/ is compact and L is a Q.g/-positive linear func-tional on RdŒ x �, then L is a K.f/-moment functional.

We briefly discuss Theorem 12.27. If k D 1, then Q.f/ D T.f/. However, fork D 2,

Q.f/ DX

RdŒ x �2 C f1

XRdŒ x �

2 C f2X

RdŒ x �2;


so Q.f/ differs from the preordering T.f/ by the summand f1f2P

RdŒ x �2. If k D 3,then

Q.f/ DX

RdŒ x �2Cf1

XRdŒ x �

2 C f2X

RdŒ x �2 C f3

XRdŒ x �

2 C f1f2X

RdŒ x �2 ;

that is, the sets gP

RdŒ x �2 with g D f1f3; f2f3; f1f2f3 do not enter into the definitionof Q.f/. For k D 4, no products of three or four generators appear in the definitionof Q.f/. For large k, only a small portion of mixed products occur in Q.f/ andTheorem 12.27 is an essential strengthening of Theorem 12.24.

The next corollary characterizes in terms of moment functionals when a Radonmeasure on a compact semi-algebraic set has a bounded density with respect toanother Radon measure. A version for closed sets is stated in Exercise 14.11 below.

Corollary 12.29 Suppose that the semi-algebraic set K.f/ is compact. Let � and �be finite Radon measures on K.f/ and let L� and L� be the corresponding momentfunctionals on RdŒ x �. There exists a function ' 2 L1.K.f/; �/, '.x/ � 0 �-a.e. onK.f/, such that d� D 'd� if and only if there is a constant c > 0 such that

L�.g/ � cL�.g/ for g 2 T.f/: (12.23)

Proof Choosing c � k'kL1.K.f/;�/, the necessity of (12.23) is easily verified.To prove the converse we assume that (12.23) holds. Then, by (12.23), L WD

cL� � L� is a T.f/-positive linear functional on RdŒ x � and hence a K.f/-momentfunctional by Theorem 12.25. Let � be a representing measure of L, that is, L D L� .Then we have L� C L� D cL�. Hence both � C � and c� are representing measuresof the K.f/-moment functional cL�. Since K.f/ is compact, c� is determinate byProposition 12.17, so that �C� D c�. In particular, this implies that � is absolutelycontinuous with respect to�. Therefore, by the Radon–Nikodym theorem A.3, d� D'd� for some function ' 2 L1.K.f/; �/, '.x/ � 0 �-a.e. on K.f/. Since �C� D c�,for each Borel subset M of K.f/ we have

�.M/ D c�.M/ � �.M/ DZM.c � '.x//d� � 0:

Therefore, c�'.x/ � 0 �-a.e., so that ' 2 L1.K.f/; �/ and k'kL1.K.f/;�/ � c: utWe close this section by restating Theorems 12.24 and 12.25 in the special case

of compact real algebraic sets.

Corollary 12.30 Suppose that I is an ideal of RdŒ x � such that the real algebraicset V WD Z.I/ D fx 2 Rd W f .x/ D 0 for f 2 Ig is compact.

(i) If h 2 RdŒ x � satisfies h.x/ > 0 for all x 2 V, then h 2PRdŒ x �2 C I.(ii) If p 2 RdŒ x �=I and p.x/ > 0 for all x 2 V, then p 2P.RdŒ x �=I/2.

(iii) If q 2 RŒV� � RdŒ x �= OI and q.x/ > 0 for all x 2 V, then q 2PRŒV�2.(iv) Each positive linear functional on RdŒ x � which annihilates I is a V-moment

functional.


Proof Put f1 D 1; f2 D h1; f3 D �h1; : : : ; f2m D hm; f2mC1 D �hm, where h1; : : : ; hmis a set of generators of I. Then, by (12.9), the preordering T.f/ is

PRdŒ x �2 C I

and the semi-algebraic set K.f/ is V D Z.I/. Therefore, Theorem 12.24 yields (i).Since I OI , (i) implies (ii) and (iii).

Clearly, a linear functional on RdŒ x � is T.f/-positive if it is positive andannihilates I. Thus (iv) follows at once from Theorem 12.25. utExample 12.31 (Moment problem on unit spheres) Let Sd�1 be the unit sphereof Rd. Then Sd�1 is the real algebraic set Z.I/ for the ideal I generated byh1.x/ D x21 C � � � C x2d � 1:

Suppose that L is a linear functional on RdŒ x � such that

L. p2/ � 0 and L..x21 C � � � C x2d � 1/p/ D 0 for p 2 RdŒ x �:

Then it follows from Corollary 12.30(iv) that L is an Sd�1-moment functional.Further, if q 2 RŒSd�1� is strictly positive on Sd�1; that is, q.x/ > 0 for x 2 Sd�1,

then q 2PRŒSd�1�2 by Corollary 12.30(iii). ı

12.4 The Representation Theorem for Archimedean Modules

The main aim of this section is to derive the representation theorem for Archimedeanquadratic modules (Theorem 12.35) and its application to the moment problem(Theorem 12.36). Our proof is a combination of functional-analytic and algebraicmethods, but we avoid the use of Hilbert space operators! At the end of the nextsection we give elegant and extremely short proofs of these results based on Hilbertspace methods. In view of an application given in Sect. 12.6 we also prove therepresentation theorem for Archimedean semirings.

Let E be a real vector space and let C be a cone in E, that is, C is a subset ofE satisfying a C b 2 C and �a 2 C for a; b 2 C and � > 0. The cone yields anordering “ �00 on E defined by x � y if and only if y � x 2 C.

An element e 2 C is called an order unit if, given a 2 E, there exists � > 0 suchthat �e�a 2 C. Since this also holds for �a, there is a Q� > 0 such that Q� e˙a 2 C.

Suppose that e 2 C is an order unit. For a 2 E we define

kake WD inf f� > 0 W ��e � a � �eg and q.a/ D inf f� > 0 W a � �eg:

Then k � ke is a seminorm and q is a sublinear functional on E. The latter means thatq.�a/ D �q.a/ and q.aC b/ � q.a/C q.b/ for a; b 2 E and � � 0. Note that q isthe Minkowski functional of the convex set e � C, that is,

q.a/ D inf f� > 0 W a 2 �.e � C/g:

By definition we have kake D max.q.a/; q.�a// for a 2 E.

12.4 The Representation Theorem for Archimedean Modules 299

Now we introduce some terminology adapted from operator algebras. Let C0denote the set of linear functionals ' on E which are C-positive, that is, '.c/ � 0for all c 2 C. The elements of C0

e WD f' 2 C0 W '.e/ D 1g are called C-states andan extreme point of the convex set C0

e is called a pure C-state of E.First we prove the following sharpening of Proposition 12.11.

Proposition 12.32 Suppose that e 2 C is an order unit. If a0 2 E and a0 … C, thenthere exists a pure C-state ' such that '.e � a0/ D q.e � a0/ and '.a0/ � 0.Proof Since a0 … C, we have e�a0 … e�C and hence q.e�a0/ � 1 by the definitionof the Minkowski functional q of e�C. Hence '.e/ D 1 and '.e� a0/ D q.e� a0/imply that '.a0/ � 0.

Let E0 denote the real vector space of all linear functionals on E and � the weaktopology �.E0;E/ on E0. In this proof we essentially use the sets

F WD f' 2 E0 W '.a/ � q.a/ for a 2 Eg; F0 WD f' 2 F W '.e � a0/ D q.e � a0/g:

First we verify the inclusions

F0 C0e F: (12.24)

Let ' 2 C0. If a 2 E and a � �e, then '.a/ � �'.e/ for � > 0 and therefore

'.a/ � q.a/'.e/ (12.25)

by the definition of q. Hence ' 2 F if ' 2 C0e. This gives the second inclusion.

Now let ' 2 F0. Then, for c 2 C we have �c � 0 � �e for all � > 0, sothat q.�c/ D 0. Therefore, �'.c/ D '.�c/ � q.�c/ D 0, that is, '.c/ � 0.Thus, ' 2 C0 and the inequality (12.25) applies. By ' 2 F0 and (12.25), we have'.e/ � q.e/ � 1 and q.e � a0/ D '.e � a0/ � q.e � a0/'.e/: Since q.e � a0/ � 1as noted above, we get '.e/ D 1, that is, ' 2 C0

e. This proves the first inclusion of(12.24).

Next we study the set F0. Clearly, '0.˛.e � a0// WD ˛q.e � a0/, ˛ 2 R, definesa linear functional '0 on the vector space E0 WD R � .e� a0/ such that '0.b/ � q.b/for b 2 E0. By the Hahn–Banach theorem '0 extends to a linear functional ' on Esatisfying '.a/ � q.a/ for a 2 E. Since '.e � a0/ D '0.e � a0/ D q.e � a0/, 'belongs to F0. Thus F0 is not empty. Obviously, F0 is convex.

Now we show that F0 is �-compact. Let U D fa 2 E W kake � 1g. Recall thatthe polar Uı of the set U is defined by Uı D f' 2 E0 W j'.a/j � 1 for a 2 Ug.By the Alaoglu–Bourbaki theorem (see e.g. [Ru1, Ch. 3, Theorem 3.15]), the polarUı is �-compact. Clearly, Uı D f' 2 E0 W j'.x/j � kake; a 2 Eg. If ' 2 F0, then'.a/ � q.a/ � kake and �'.a/ � q.�a/ � kake, so j'.a/j � kake for a 2 E, thatis, ' 2 Uı. Thus F0 Uı. Since F0 is obviously �-closed in Uı, F0 is �-compact.

Thus we proved that F0 is a nonempty �-compact convex subset of E0. By theKrein–Milman theorem ([Ru1, Ch. 3, Theorem 3.22]), F0 has an extreme point, say


'. We show that ' is an extreme point of the larger set F. For let ' D 12.'1 C '2/,

where '1; '2 2 F. From q.e � a0/ D '.e � a0/ D 12.'1.e � a0/ C '2.e � a0// �

q.e � a0/ we conclude that 'j.e � a0/ D q.e � a0/, so 'j 2 F0 for j D 1; 2. Hence'1 D '2 D ', so ' is an extreme point of F. Therefore, since ' 2 F0 C0

e andC0e F by (12.24), ' is also an extreme point of C0

e, that is, ' is a pure C-state. utProposition 12.33 Let Q be an Archimedean quadratic module or an Archimedeansemiring of a unital commutative real algebra A. Then each pure Q-state ' is acharacter, that is,

'.ab/ D '.a/'.b/ for a; b 2 A: (12.26)

In the proof of this proposition we use the following technical lemma.

Lemma 12.34 Let Q be an Archimedean quadratic module of A and let � > > 0.Suppose that a is an element of A such that aQ Q and � a 2 Q. If ' is aQ-positive linear functional on A, then '..� � a/c/ � 0 for c 2 Q.

Proof Without loss of generality we can assume that � D 1 and '.1/ D 1. Considerthe Taylor polynomial pn.t/ of the function

p1 � t , that is,

pn.t/ DnX

kD0.�1/k

1=2

k

!tk;

and the polynomial qn.t/ WD pn.t/2 � .1 � t/ DPk �nktk. We compute

�nk D .�1/knX

jDk�n

1=2

j

! 1=2

k � j

!

for n < k < 2n and �nk D 0 otherwise. Since the term for the index j in the sum hassign .�1/j�1.�1/k�j�1 D .�1/k, we have �nk � 0 for all n and k.

Since � a 2 Q and aQ Q by assumption, it follows from the identity

k � ak Dk�1XjD0

k�1�ja j. � a/

that k � ak 2 Q and hence ak � k for all k 2 N.Let c 2 Q. We choose ˛ > 0 such that ˛ � c 2 Q. Since aQ Q and qn.t/ has

only nonnegative coefficients �nk, we get qn.a/Q Q. Therefore, qn.a/.˛ � c/ 2 Qand so '.qn.a/.˛ � c// � 0. Hence we derive

'.qn.a/c/ � ˛'.qn.a// D ˛nX

kD0�nk'.a

k/ � ˛nX

kD0�nk

k D ˛qn./: (12.27)

12.4 The Representation Theorem for Archimedean Modules 301

Since Q is a quadratic module and ' is Q-positive, we have '. pn.a/2c/ � 0: Usingthis fact, the identity qn.a/ D pn.a/2 � .1 � a/; and (12.27) we obtain

'..1 � a/c/ D '. pn.a/2c/� '.qn.a/c/ � �'.qn.a/c/ � �˛qn./: (12.28)

Since 0 < < 1, we have pn./ ! p1 � and hence qn./ ! 0. Therefore,letting n!1 in (12.28) we get '..1 � a/c/ � 0. utProof of Proposition 12.33 First we suppose that Q is an Archimedean quadraticmodule. From the identity 4a D .aC1/2�.a�1/2, a 2 A, it follows that A D Q�Q.Hence it suffices to prove (12.26) for squares a D c2 and b D d2, where c; d 2 A.Clearly, '.c2/ � 0, since c2 2 Q.

Case 1: '.c2/ D 0. There is a � > 0 such that � � d2 2 Q. Since Q is aquadratic module and ' is Q-positive, c2.� � d2/ 2 Q, so that 0 � c2d2 � �c2 andhence 0 � '.c2d2/ � �'.c2/ D 0. Thus, '.c2d2/ D 0 and (12.26) holds for a D c2

and b D d2.Case 2: '.c2/ > 0. Clearly, '1.�/ D '.c2/�1'.c2 �/ is a Q-positive state. Let us

choose � > 0 such that �2� c2 2 Q. Since ' is Q-positive, '.c2/ � �

2< � and

hence '.� � c2/ > 0. Define '2.�/ WD '.� � c2/�1'..� � c2/ �/. Since �2� c2 2 Q,

it follows from Lemma 12.34 that the functional '..� � c2/ �/ is Q-positive. Hence'2 is a Q-state. By construction the pure Q-state ' is the convex combination

' D ��1'.c2/ '1 C ��1'.� � c2/ '2

of the two Q-states '1 and '2. Hence '1 D '. This yields '.c2b/ D '.c2/'.b/ forb 2 A which proves (12.26) when '.c2/ > 0.

Now we assume that Q is an Archimedean semiring. In this case, Lemma 12.34is not needed; the proof is very similar and even simpler. For a 2 A, there exists a� > 0 such that �Ca 2 Q, so that a D .�Ca/�� 2 Q�Q. Thus, A D Q�Q. Henceit suffices to verify (12.26) for a; b 2 Q: Then '.a/ � 0, since ' is Q-positive.

Case 1: '.a/ D 0.Let b 2 Q and choose � > 0 such that ��b 2 Q. Then .��b/a 2 Q and ab 2 Q

(because Q is a semiring!), so that '..� � b/a/ D �'.a/ � '.ab/ D �'.ab/ � 0and '.ab/ � 0. Hence '.ab/ D 0, so that (12.26) holds.

Case 2: '.a/ > 0.We choose � > 0 such that .��a/ 2 Q and '.��a/ > 0. Since Q is a semiring,

'1.�/ WD '.a/�1'.a�/ and '2.�/ WD '.� � a/�1'..� � a/�/ are Q-states satisfying' D ��1'.a/ '1 C ��1'.� � a/ '2: The latter is a convex combination of two Q-states. Since ' is a pure Q-state, '1 D '. Hence '.ab/ D '.a/'.b/ for b 2 A. ut

The following theorem is the representation theorem for Archimedean quadraticmodules and semirings. It has been discovered, in various versions, by a number ofauthors including M.H. Stone, R.V. Kadison, J.-L. Krivine, T. Jacobi, and others.


Theorem 12.35 Suppose that Q is an Archimedean quadratic module or anArchimedean semiring of a unital commutative real algebra A. Let a0 2 A. If'.a0/ > 0 for all ' 2 K.Q/, then a0 2 Q.

Proof Assume to the contrary that a0 … Q. Then, by Proposition 12.32, there is apure Q-state ' such that '.a0/ � 0. Since ' is a character by Proposition 12.33,' 2 K.Q/. Then '.a0/ � 0 contradicts the assumption. ut

For our treatment of the moment problem we will need the following applicationof Theorem 12.35. Assertion (i) is usually called the Archimedean Positivstellensatz.

Theorem 12.36 Let f D f f1; : : : ; fkg be a finite subset of RdŒ x �. Suppose that thequadratic module Q.f/ defined by (12.4) is Archimedean.

(i) If h 2 RdŒ x � satisfies f .x/ > 0 for all x 2 K.f/, then h 2 Q.f/:(ii) Any Q.f/-positive linear functional L on RdŒ x � is a K.f/-moment functional,

that is, there exists a measure � 2 MC.Rd/ supported on the compact set K.f/such that L. f / D R f .x/ d�.x/ for f 2 RdŒ x �.

Proof

(i) Set A D RdŒ x � and Q D Q.f/. As noted in Sect. 12.1, characters � of Acorrespond to points x� Š .�.x1/; : : : ; �.xd// of Rd and K.Q/ D K.f/. Hencethe assertion follows at once from Theorem 12.35.

(ii) Combine (i) and Haviland’s Theorem 1.12 (ii)!(iv). ut

12.5 The Operator-Theoretic Approach to the MomentProblem

The spectral theory of self-adjoint operators in Hilbert space is well suited to themoment problem and provides powerful techniques for the study of this problem.The technical tool that relates the multidimensional moment problem to Hilbertspace operator theory is the Gelfand–Naimark–Segal construction, briefly the GNS-construction. We develop this construction first for a general �-algebra (see also[Sm4, Section 8.6]) and then we specialize to the polynomial algebra.

Suppose that A is a unital (real or complex) �-algebra. Let K D R or K D C.

Definition 12.37 Let .D; h�; �i/ be a unitary space. A �-representation of A on.D; h�; �i/ is an algebra homomorphism � of A into the algebra L.D/ of linearoperators mapping D into itself such that �.1/' D ' for ' 2 D and

h�.a/'; i D h'; �.a�/ i for a 2 A; '; 2 D: (12.29)

The unitary space D is called the domain of � and denoted by D.�/. A vector' 2 D is called algebraically cyclic, briefly a-cyclic, for � if D D �.A/'.

12.5 The Operator-Theoretic Approach to the Moment Problem 303

Suppose that L is a positive linear functional on A, that is, L is a linear functionalsuch that L.a�a/ � 0 for a 2 A. Then, by Lemma 2.3, the Cauchy–Schwarzinequality holds:

jL.a�b/j2 � L.a�a/L.b�b/ for a; b 2 A: (12.30)

Lemma 12.38 NL WD fa 2 A W L.a�a/ D 0g is a left ideal of the algebra A.

Proof Let a; b 2 NL and x 2 A. Using (12.30) we obtain

jL..xa/�xa/j2 D jL..x�xa/�a/j2 � L..x�xa/�x�xa/L.a�a/ D 0;

so that xa 2 NL. Applying again (12.30) we get L.a�b/ D L.b�a/ D 0. Hence

L..aC b/�.aC b// D L.a�a/C L.b�b/C L.a�b/C L.b�a/ D 0;

so that aC b 2 NL. Obviously, �a 2 NL for � 2 K. utHence there exist a well-defined scalar product h�; �iL on the quotient vector space

DLDA=NL and a well-defined algebra homomorphism �L W A!L.DL/ given by

haCNL; bCNLiL D L.b�a/; �L.a/.bCNL/ D abCNL; a; b 2 A: (12.31)

Let HL denote the Hilbert space completion of the pre-Hilbert space DL. If noconfusion can arise we write h�; �i for h�; �iL and a for a C NL. Then we have�L.a/b D ab, in particular �L.1/a D a, and

h�L.a/b; ci D L.c�ab/ D L..a�c/�b/ D hb; �L.a�/ci; a; b; c 2 A: (12.32)

Clearly, DL D �L.A/1. Thus, we have shown that �L is a �-representation of A onthe domain D.�L/ D DL and 1 is an a-cyclic vector for �L. Further, we have

L.a/ D h�L.a/1; 1i for a 2 A: (12.33)

Definition 12.39 �L is called the GNS-representation of A associated with L.

We show that the GNS-representation is unique up to unitary equivalence. Let �be another �-representation of A with a-cyclic vector ' 2 D.�/ on a dense domainD.�/ of a Hilbert space G such that L.a/ D h�.a/'; 'i for all a 2 A. For a 2 A,

k�.a/'k2 D h�.a/'; �.a/'i D h�.a�a/'; 'i D L.a�a/

and similarly k�L.a/1k2 D L.a�a/. Hence there is an isometric linear map U givenby U.�.a/'/ D �L.a/1; a 2 A; of D.�/ D �.A/' onto D.�L/ D �L.A/1. Since


the domains D.�/ and D.�L/ are dense in G and HL, respectively, U extends bycontinuity to a unitary operator of G onto HL. For a; b 2 A we derive

U�.a/U�1.�L.b/1/ D U�.a/�.b/' D U�.ab/' D �L.ab/1 D �L.a/.�L.b/1/;

that is, U�.a/U�1' D �L.a/' for ' 2 D.�L/ and a 2 A. By definition, this meansthat the �-representations � and �L are unitarily equivalent.

Now we specialize the preceding to the �-algebra CdŒ x � � CŒx1; : : : ; xd� withinvolution determined by .xj/� WD xj for j D 1; : : : ; d.

Suppose that L is a positive linear functional on CdŒ x �. Since .xj/� D xj, itfollows from (12.32) that Xj WD �L.xj/ is a symmetric operator on the domainDL. The operators Xj and Xk commute (because xj and xk commute in CdŒ x �)and Xj leaves the domain DL invariant (because xjCdŒ x � CdŒ x �). That is,.X1; : : : ;Xd/ is a d-tuple of pairwise commuting symmetric operators acting onthe dense invariant domain DL D �L.CdŒ x �/1 of the Hilbert space HL. Notethat this d-tuple .X1; : : : ;Xd/ essentially depends on the given positive linearfunctional L.

The next theorem is the crucial result of the operator approach to the multidi-mensional moment problem and it is the counterpart of Theorem 6.1. It relatessolutions of the moment problem to spectral measures of strongly commutingd-tuples .A1; : : : ;Ad/ of self-adjoint operators which extend our given d-tuple.X1; : : : ;Xd/.

Theorem 12.40 A positive linear functional L on the �-algebraCdŒ x � is a momentfunctional if and only if there exists a d-tuple .A1; : : : ;Ad/ of strongly commutingself-adjoint operators A1; : : : ;Ad acting on a Hilbert space K such that HL is asubspace of K and X1 A1; : : : ;Xd Ad. If this is fulfilled and E.A1;:::;Ad/ denotesthe spectral measure of the d-tuple .A1; : : : ;Ad/, then �.�/ D hE.A1;:::;Ad/.�/1; 1iKis a solution of the moment problem for L.

Each solution of the moment problem for L is of this form.

First we explain the notions occurring in this theorem (see [Sm9, Chapter 5] forthe corresponding results and more details).

A d-tuple .A1; : : : ;Ad/ of self-adjoint operators A1; : : : ;Ad acting on a Hilbertspace K is called strongly commuting if for all k; l D 1; : : : ; d; k ¤ l; the resolvents.Ak � iI/�1 and .Al � iI/�1 commute, or equivalently, the spectral measures EAk andEAl commute (that is, EAk.M/EAl.N/ D EAl.N/EAk.M/ for all Borel subsets M;Nof R). (If the self-adjoint operators are bounded, strong commutativity and “usual”commutativity are equivalent.) The spectral theorem states that, for such a d-tuple,there exists a unique spectral measure E.A1;:::;Ad/ on the Borel �-algebra of Rd suchthat

Aj DZRd�j dE.A1;:::;Ad/.�1; : : : ; �d/; j D 1; : : : ; d:

12.5 The Operator-Theoretic Approach to the Moment Problem 305

The spectral measure E.A1;:::;Ad/ is the product of spectral measures EA1 ; � � �EAd .Therefore, if M1; : : : ;Md are Borel subsets of R, then

E.A1;:::;Ad/.M1 � � � � �Md/ D EA1 .M1/ � � �EAd.Md/: (12.34)

Proof of Theorem 12.40 First assume that L is the moment functional and let � be arepresenting measure of L. It is well-known and easily checked by the precedingremarks that the multiplication operators Ak, k D 1; : : : ; d, by the coordinatefunctions xk form a d-tuple of strongly commuting self-adjoint operators on theHilbert space K WD L2.Rd; �/ such that HL K and Xk Ak for k D 1; : : : ; d.The spectral measure E WD E.A1;:::;Ad/ of this d-tuple acts by E.M/f D �M � f ,f 2 L2.Rd; �/, where �M is the characteristic function of the Borel set M Rd.This implies that hE.M/1; 1iK D �.M/. Thus, �.�/ D hE.�/1; 1iK.

Conversely, suppose that .A1; : : : ;Ad/ is such a d-tuple. By the multidimensionalspectral theorem [Sm9, Theorem 5.23] this d-tuple has a joint spectral measureE.A1;:::;Ad/. Put �.�/ WD hE.A1;:::;Ad/.�/1; 1iK. Let p 2 CdŒ x �. Since Xk Ak,

p.X1; : : : ;Xd/ p.A1; : : : ;Ad/:

Therefore, since the polynomial 1 belongs to the domain of p.X1; : : : ;Xd/, it is alsoin the domain of p.A1; : : : ;Ad/. Then

ZRdp.�/ d�.�/ D

ZRd

p.�/ dhE.A1;:::;Ad/.�/1; 1iK D hp.A1; : : : ;Ad/1; 1iK

D hp.X1; : : : ;Xd/1; 1i D h�L. p.x1; : : : ; xd//1; 1i D L. p.x1; : : : ; xd//;

where the second equality follows from the functional calculus and the last from(12.33). This shows that � is a solution of the moment problem for L. utProposition 12.41 Suppose Q is an Archimedean quadratic module of a commu-tative real unital algebra A. Let L0 be a Q-positive R-linear functional on A andlet �L be the GNS representation of its extension L to a C-linear functional on thecomplexification AC D AC iA. Then all operators �L.a/, a 2 AC, are bounded.

Proof SinceP.AC/

2 D PA2 by Lemma 2.17(ii) and

PA2 Q, L is a positive

linear functional on AC, so the GNS representation �L is well-defined.It sufffices to prove that �L.a/ is bounded for a 2 A. Since Q is Archimedean,

� � a2 2 Q for some � > 0. Let x 2 AC. By Lemma 2.17(ii), x�x.� � a2/ 2 Q andhence L.x�xa2/ D L0.x�xa2/ � �L0.x�x/ D �L.x�x/, since L0 is Q-positive. Then

k�L.a/�L.x/1k2 D h�L.a/�L.x/1; �L.a/�L.x/1i D h�L..ax/�ax/1; 1i

D L..ax/�ax/ D L.x�xa2/ � �L.x�x/ D �k�L.x/1k2;

where we used (12.29) and (12.33). That is, �L.a/ is bounded on D.�L/. ut


We now illustrate the power of the operator approach to moment problems bygiving short proofs of Theorems 12.35 and 12.36.

Proof of Theorem 12.35 Assume to the contrary that a0 … Q. Since Q isArchimedean, by Proposition 12.11 there is a Q-positive R-linear functional L0on A such that L0.1/ D 1 and L0.a0/ � 0. Let �L be the GNS representation of itsextension to a C-linear (positive) functional L on the unital commutative complex�-algebra AC.

Let c 2 Q. If x 2 AC, then x�xc 2 Q by Lemma 2.17(ii), so L0.x�xc/ � 0, and

h�L.c/�L.x/1; �L.x/1i D L.x�xc/ D L0.x�xc/ � 0 (12.35)

by (12.33). This shows that the operator �L.c/ is nonnegative.For a 2 AC, the operator �L.a/ is bounded by Proposition 12.41. Let �L.a/

denote its continuous extension to the Hilbert space HL. These operators form aunital commutative �-algebra of bounded operators. Its completion B is a unitalcommutative C�-algebra.

Let � be a character of B. Then Q�.�/ WD �. �L.�/ / is a character of A. Ifc 2 Q, then �L.c/ � 0 by (12.35) and so �L.c/ � 0. Hence Q� is Q-positive,that is, Q� 2 K.Q/. Therefore, Q�.a0/ D �.�L.a0/ / > 0 by the assumption ofTheorem 12.35. Therefore, if we realize B as a C�-algebra of continuous functionson a compact Hausdorff space, the function corresponding to �L.a0/ is positive, soit has a positive minimum ı. Then �L.a0/ � ı � I and hence

0 < ı D ıL.1/ D hı1; 1i � h�L.a0/1; 1i D L.a0/ D L0.a0/ � 0;

which is the desired contradiction. utProof of Theorem 12.36(ii) We extend L to a C-linear functional, denoted againby L, on CdŒ x � and consider the GNS representation �L. By Proposition 12.41,the symmetric operators �L.x1/; : : : ; �L.xd/ are bounded. Hence their continuousextensions to the whole Hilbert space HL are pairwise commuting bounded self-adjoint operators A1; : : : ;Ad. Therefore, by Theorem 12.40, if E denotes the spectralmeasure of this d-tuple .A1; : : : ;Ad/, then �.�/ D hE.�/1; 1iHL is a solution of themoment problem for L.

Since the operatorsAj are bounded, the spectral measure E, hence�, has compactsupport. (In fact, supp E Œ�kA1k; kA1k� � � � � � Œ�kAdk; kAdk�.) Hence, since Lis Q.f/-positive by assumption, Proposition 12.18 implies that supp� K.f/. Thisshows that L is a K.f/-moment functional. ut

The preceding proof of Theorem 12.36(ii) based on the spectral theorem isprobably the most elegant approach to the moment problem for Archimedeanquadratic modules. Now we derive Theorem 12.36(i) from Theorem 12.36(ii).

Proof of Theorem 12.36(i) We argue in the same manner as in the second proof ofTheorem 12.24 in Sect. 12.3. Assume to the contrary that h … Q.f/. Since Q.f/is Archimedean, Proposition 12.11 and Theorem 12.36(ii) apply to Q.f/. By these

12.6 The Moment Problem for Semi-Algebraic Sets Contained in Compact. . . 307

results, there is a Q.f/-positive linear functional L on RdŒ x � satisfying L.1/ D 1 andL.h/ � 0, and this functional is a K.f/-moment functional. Then there is a measure� 2 MC.Rd/ supported on K.f/ such that L. p/ D R

p d� for p 2 RdŒ x �. (Notethat K.f/ is compact by Corollary 12.9.) Again h.x/ > 0 on K.f/, L.1/ D 1, andL.h/ � 0 lead to a contradiction. ut

12.6 The Moment Problem for Semi-Algebraic SetsContained in Compact Polyhedra

In this section, f1; : : : ; fk are polynomials of RdŒ x � such that the first m polynomialsf1; : : : ; fm, where 1 � m � k, are linear. By a linear polynomial we mean apolynomial of degree at most one. We abbreviate

Of D f f1; : : : ; fmg; f D f f1; : : : ; fkg:

Then K.Of/ is a semi-algebraic set defined by linear polynomials f1; : : : ; fm; such aset is called a polyhedron. The general semi-algebraic set K.f/ is contained in thepolyhedron K.Of/:Definition 12.42 Let P.f/ denote the semiring generated by f1; : : : ; fk, that is, P.f/consists of all finite sums of terms of the form

˛ f n11 � � � f nkk ; where ˛ � 0; n1; : : : ; nk 2 N0: (12.36)

Note that P.f/ is not a quadratic module in general.The following lemma goes back to H. Minkowski. In the optimization literature

it is called Farkas’ lemma. We will use it in the proof of Theorem 12.44 below.

Lemma 12.43 Let h; f1; : : : ; fm be linear polynomials of RdŒ x � such that the setK.Of/ is not empty. If h.x/ � 0 on K.Of/, there exist numbers �0 � 0; : : : ; �m � 0 suchthat h D �0 C �1f1 C � � � C �mfm:Proof Let E be the vector space spanned by the polynomials 1; x1; : : : ; xd and C thecone in E generated by 1; f1; : : : ; fm. It is easily shown that C is closed in E.

We have to prove that h 2 C. Assume to the contrary that g … C. Then, bythe separation of convex sets (Theorem A.26(ii)), there exists a C-positive linearfunctional L on E such that L.h/ < 0. In particular, L.1/ � 0, because 1 2 C.

Without loss of generality we can assume that L.1/ > 0. Indeed, if L.1/ D 0,we take a point x0 of the nonempty (!) set K. Of / and replace L by L0 D L C "lx0 ,where lx0 denotes the point evaluation at x0 on E. Then L0 is C-positive as well andL0.h/ < 0 for small " > 0.

Define a point x WD L.1/�1.L.x1/; : : : ;L.xd// 2 Rd. Then L.1/�1L is theevaluation lx at the point x for the polynomials x1; : : : ; xd and for 1, hence on the


whole vector space E. Therefore, fj.x/ D lx. fj/ D L.1/�1L. fj/ � 0 for all j, so thatx 2 K. Of /, and g.x/Dlx.h/DL.1/�1L.h/ < 0. This contradicts the assumption. utTheorem 12.44 Let f1; : : : ; fk be polynomials of RdŒ x � such that the polynomialsf1; : : : ; fm; 1 � m � k, are linear. Suppose that the polyhedron K. Of / is compactand nonempty. Let h 2 RdŒ x �. If h.x/ > 0 for all x 2 K.f/, then h 2 P.f/.

Proof First we show that the semiring P.f/ is Archimedean. Let i 2 f1; : : : ; dg.Since the set K. Of / is compact, there exists a � > 0 such that � ˙ xi � 0 onK. Of /. Hence, since K. Of / is nonempty, Lemma 12.43 implies that .�˙ xi/ 2 P.f/.Therefore, P.f/ is Archimedean by Lemma 12.7(ii).

Now we apply Theorem 12.35 to the Archimedean semiring Q D P.f/ of thealgebra A D RdŒ x �. Clearly, each character � of A is given by a point x 2 Rd. If �is P.f/-positive, then �. fj/ D fj.x/ � 0 for all j D 1; : : : ; k and therefore x 2 K.f/.Thus, K.Q/ D K.f/ and Theorem 12.35 yields the assertion. utThe following theorem is the main result of this section on the moment problem.

Theorem 12.45 Retain the assumptions and the notation of Theorem 12.44. Let Lbe a linear functional on RdŒ x �. Then L is a K.f/-moment functional if and only if

L.f n11 � � � f nkk / � 0 for all n1; : : : ; nk 2 N0: (12.37)

Proof The only if part is obvious. Conversely, suppose that (12.37) is satisfied. Bythe definition of the semiring P.f/, (12.37) means that L is P.f/-positive. Hence, byTheorem 12.44 and Haviland’s Theorem 1.12, L is a K.f/-moment functional. ut

Let us consider the important special case m D k. Suppose that K.f/ isa nonempty compact set. Since k D m, all polynomials f1; : : : ; fk are linear.Hence K.f/ D K. Of / is a polyhedron. Then, by Theorem 12.45, (12.37) is asolvability condition for the moment problem of the compact polyhedron K.f/ andTheorem 12.44 gives a description of strictly positive polynomials on K.f/.

12.7 Examples and Applications

Throughout this section, f D f f1; : : : ; fkg is a finite subset of RdŒ x � and L denotes alinear functional on RdŒ x �:

If L is a K.f/-moment functional, it is obviously T.f/-positive, Q.f/-positive, andP.f/-positive. Theorems 12.25, 12.36, and 12.45 deal with the converse implicationand are the main solvability criteria for the moment problem proved in this chapter.

First we discuss Theorems 12.25 and 12.36(ii). Theorem 12.25 applies to eachcompact semi-algebraic set K.f/ and implies that L is a K.f/-moment functional ifand only if it is T.f/-positive. For Theorem 12.36(ii) the compactness of the set K.f/is not sufficient; it requires that the quadratic module Q.f/ is Archimedean. In thiscase, L is a K.f/-moment functional if and only if it is Q.f/-positive.

12.7 Examples and Applications 309

Example 12.46 Let us begin with a single polynomial f 2 RdŒ x � for which the setK. f / D fx 2 Rd W f .x/ � 0g is compact. (A simple example is the d-ellipsoidgiven by f .x/ D 1� a1x21 � � � � � adx2d, where a1 > 0; : : : ; ad > 0.) Clearly, T. f / DQ. f /. Then, L is a K. f /-moment functional if and only if it is T. f /-positive, orequivalently, if L and Lf are positive functionals on RdŒ x �.

Now we add further polynomials f2; : : : ; fk and set f D f f ; f2; : : : ; fkg. (Forinstance, one may take coordinate functions as fj D xl.) Since T. f / is Archimedean(by Proposition 12.22, because K. f / is compact), so is the quadratic module Q.f/.Therefore, L is a K.f/-moment functional if and only if it is Q. f /-positive, orequivalently, if L;Lf ;Lf2 ; : : : ;Lfk are positive functionals on RdŒ x �. ıExample 12.47 (d-dimensional compact interval Œa1; b1�� Œad; bd�) Let aj; bj 2R, aj < bj; and set f2j�1 WD bj � xj, f2j WD xj � aj; for j D 1; : : : ; d. Then the semi-algebraic set K.f/ for f WD f f1; : : : ; f2dg is the d-dimensional interval Œa1; b1�� Œad; bd�.

Put �j D jajj C jbjj: Then �j � xj D f2j�1 C �j � bj and �j C xj D f2j C �j C ajare Q.f/, so each xj is a bounded element with respect to the quadratic module Q.f/.Hence Q.f/ is Archimedean by Lemma 12.7(ii).

Thus, L is a K.f/-moment functional if and only if it is Q. f /-positive, orequivalently, if Lf1 ;Lf2 ; : : : ;Lfk are positive functionals, that is,

L..bj�xj/p2/ � 0 and L..xj�aj/p2/ � 0 for j D 1; : : : ; d; p 2 RdŒ x �:(12.38)

Clearly, (12.38) implies that L itself is positive, since L D .b1�a1/�1.Lf1CLf2 /. ıExample 12.48 (1-dimensional interval Œa; b�) Let a < b, a; b 2 R and let l; n 2 N

be odd. We set f .x/ WD .b� x/l.x�a/n. Then K.f / D Œa; b� and T.f / DPRŒx�2CfP

RŒx�2. Hence, by Theorem 12.25, a linear functional L on RŒx� is an Œa; b�-moment functional if and only if L and Lf are positive functionals on RŒx�.

This result extends Hausdorff’s Theorem 3.13. It should be noted that thissolvability criterion holds for arbitrary (!) odd numbers l and n, while the equalityPos.Œa; b�/ D T. f / is only true if l D n D 1, see Exercise 3.4 b. in Chap. 3. ıExample 12.49 (Simplex in Rd; d � 2) Let f1 D x1; : : : ; fd D xd; fdC1 D1 �Pd

iD1 xi; k D dC 1. Clearly, K.f/ is the simplex

Kd D fx 2 Rd W x1 � 0; : : : ; xd � 0; x1 C � � � C xd � 1 g:

Note that 1 � xj D fdC1 C Pi¤j fi and 1 C xj D 1 C fj. Therefore, 1 ˙ xj 2

Q.f/ and 1˙ xj 2 P.f/. Hence, by Lemma 12.7(ii), the quadratic module Q.f/ andthe semiring P.f/ are Archimedean. Therefore, Theorem 12.36 applies to Q.f/ andTheorems 12.44 and 12.45 apply to P.f/. We restate only the results on the momentproblem.

By Theorems 12.36(ii) and 12.45, L is a Kd–moment functional if and only if

L.xip2/ � 0; i D 1; � � � ; d; and L..1� .x1 C x2 C � � � C xd//p

2/ � 0 for p 2 RdŒ x �;


or equivalently,

L.xn11 : : : xndd .1 � .x1 C � � � C xd//

ndC1 / � 0 for n1; : : : ; ndC1 2 N0: ut ı

Example 12.50 (Standard simplex �d in Rd) Let f1 D x1; : : : ; fd D xd; fdC1 D1 �Pd

iD1 xi; fdC2 D �fdC1; k D d C 2: Then the semi-algebraic set K.f/ is thestandard simplex

�d D fx 2 Rd W x1 � 0; : : : ; xd � 0; x1 C � � � C xd D 1g:

Let P0 denote the polynomials of RdŒ x � with nonnegative coefficients and I theideal generated by 1 � .x1 C � � � C xd/. Then P WD P0 C I is a semiring of RdŒ x �.Since 1˙ xj 2 P , P is Archimedean. The characters of RdŒ x � are the evaluations atpoints of Rd. Obviously, x 2 Rd gives a P-positive character if and only if x 2 �d.

Let f 2 RdŒ x � be such that f .x/ > 0 on �d. Then, f 2 P by Theorem 12.35, so

f .x/ D g.x/C h.x/.1� .x1 C � � � C xd//; where g 2 P0; h 2 RdŒ x �: (12.39)

From Theorem 12.45 it follows that L is a �d-moment functional if and only if

L.xn11 : : : xndd / � 0; L.xn11 : : : xndd .1�.x1C : : :Cxd//r/ D 0; n1; : : : ; nd 2 N0; r 2 N: ı

From the preceding example it is only a small step to derive an elegant proof ofthe following classical theorem of G. Polya.

Proposition 12.51 Suppose that f 2 RdŒ x � is a homogeneous polynomial such thatf .x/ > 0 for all x 2 Rdnf0g, x1 � 0; : : : ; xd � 0. Then there exists an n 2 N suchthat all coefficients of the polynomial .x1 C � � � C xd/nf .x/ are nonnegative.

Proof We use Example 12.50. As noted therein, Theorem 12.35 implies that fis of the form (12.39). We replace in (12.39) each variable xj; j D 1; : : : ; d; byxj.Pd

iD1 xi/�1. Since .1 �Pj xj.P

i xi/�1/ D 1 � 1 D 0; the second summand in

(12.39) vanishes after this substitution. Hence, because f is homogeneous, (12.39)yields

� Xixi��m

f .x/ D g�x1� X

ixi��1; : : : ; xd

� Xixi��1�

; (12.40)

where m D deg. f /. Since g 2 P0, g.x/ has only nonnegative coefficients.Therefore, after multiplying (12.40) by .

Pi xi/

nCm with n sufficiently large to clearthe denominators, we obtain the assertion. ut

Now we treat two examples which are applications of Theorem 12.45.

Example 12.52 Œ�1; 1�dLet k D m D 2d and f1 D 1 � x1; f2 D 1C x1; : : : ; f2d�1 D 1 � xd; f2d D 1C xd:

Then K. Of / D K. f / D Œ�1; 1�d. Therefore, by Theorem 12.45, a linear functional

12.8 Exercises 311

L on RdŒxd� is a Œ�1; 1�d-moment functional if and only if

L..1 � x1/n1 .1C x1/

n2 � � � .1 � xd/n2d�1 .1C xd/

n2d/ � 0 for n1; : : : ; n2d 2 N0: ı

Example 12.53 (Multidimensional Hausdorff moment problem on Œ0; 1�d) Set f1 Dx1; f2 D 1 � x1; : : : ; f2d�1 D xd; f2d D 1 � xd; k D 2d. Then K. Of / D Œ0; 1�d. Lets D .sn/n2Nd

0be a multisequence. We define the shift Ej of the j-th index by

.Ejs/m D s.m1;:::;mj�1;mjC1;mjC1;:::;md/; m 2 Nd0:

Proposition 12.54 The following five statements are equivalent:

(i) s is a Hausdorff moment sequence on Œ0; 1�d.(ii) Ls is a Œ�1; 1�d-moment functional on RdŒ x �:

(iii) Ls.xm11 .1 � x1/n1 � � � xmd

d .1 � xd/nd / � 0 for all n;m 2 Nd0.

(iv) ..I � E1/n1 : : : .I � Ed/nd s/m � 0 for all n;m 2 Nd

0.(v) X

j2Nd0 ;j�n

.�1/jjj n1j1

!� � � ndjd

!smCj � 0

for all n;m 2 Nd0. Here jjj WD j1 C � � � C jd and j � n means that ji � ni for

i D 1; : : : ; d.Proof (i)$(ii) holds by definition. Theorem 12.45 yields (ii)$(iii). Let n;m 2 Nd

0.We repeat the computation from the proof of Theorem 3.15 and derive

Ls.xm11 .1 � x1/

n1 � � � xmdd .1 � xd/

nd / D ..I � E1/n1 : : : .I � Ed/

nd s/m

DX

j2Nd0 ;j�n

.�1/jjj n1j1

!� � � ndjd

!smCj:

This identity implies the equivalence of conditions (iii)–(v). ut ı

12.8 Exercises

1. Suppose that Q is a quadratic module of a commutative real algebra A. Show thatQ \ .�Q/ is an ideal of A. This ideal is called the support ideal of Q.

2. Let K be a closed subset of Rd. Show that Pos.K/ is saturated.3. Formulate solvability criteria in terms of localized functionals and in terms of

d-sequences for the following sets.


a. Unit ball of Rd.b. fx 2 Rd W x21 C � � � C x2d � r2; x1 � 0; : : : ; xd � 0g.c. f.x1; x2; x3; x4/ 2 R4 W x21 C x22 � 1; x23 C x24 � 1g.d. f.x1; x2; x3/ 2 R3 W x21 C x22 C x23 � 1; x1 C x2 C x3 � 1g:e. fx 2 R2d W x21 C x22 D 1; : : : ; x22d�1 C x22d D 1g.

4. Decide whether or not the following quadratic modules Q.f/ are Archimedean.

a. f1 D x1; f2 D x2; f3 D 1 � x1x2; f4 D 4 � x1x2.b. f1 D x1; f2 D x2; f3 D 1 � x1 � x2:c. f1 D x1; f2 D x2; f3 D 1 � x1x2.

5. Let f1; : : : ; fk; g1; : : : ; gl 2 RdŒ x �. Set g D . f1; : : : ; fk; g1; : : : ; gl/, f D. f1; : : : ; fk/. Suppose that Q.f/ is Archimedean. Show that each Q.g/-positivelinear functional L is a determinate K.g/-moment functional.

6. Formulate solvability criteria for the moment problem of the following semi-algebraic sets K.f/.

a. f1 D x21 C � � � C x2d; f2 D x1; : : : ; fk D xk�1, where 2 � k � dC 1.b. f1 D x1; f2 D 2 � x1; f3 D x2; f4 D 2 � x2; f5 D x21 � x2; where d D 2.c. f1 D x21 C x22; f2 D ax1 C bx2; f3 D x2; where d D 2; a; b 2 R.

7. Let d D 2, f1 D 1�x1; f2 D 1Cx1; f3 D 1�x2; f4 D 1Cx2; f5 D 1�x21�x22 and f D. f1; f2; f3; f4; f5/: Describe the set K. f / and use Theorem 12.45 to characterizeK. f /-moment functionals.

8. Find a d-dimensional version of Exercise 7, where d � 3:9. (Reznick’s theorem [Re2])

Let f 2 RdŒ x � be a homogeneous polynomial such that f .x/ > 0 for allx 2 Rd, x ¤ 0. Prove that there exists an n 2 N such that

.x21 C � � � C x2d/nf .x/ 2

XRdŒ x �

2:

Hint: Mimic the proof of Proposition 12.51: Let T denote the preorderingPRdŒ x �2 C I, where I is the ideal generated by 1 � .x21 C � � � C x2d/. Show

that T-positive characters corresponds to points of the unit sphere, substitutexj.P

i x2i /

�1 for xj, apply Theorem 12.44 to T, and clear denominators.

12.9 Notes

The interplay between real algebraic geometry and the moment problem forcompact semi-algebraic sets and the corresponding Theorems 12.24 and 12.25 werediscovered by the author in [Sm6]. A small gap in the proof of [Sm6, Corollary 3](observed by A. Prestel) was immediately repaired by the reasoning of the aboveproof of Proposition 12.22 (taken from [Sm8, Proposition 18]).

12.9 Notes 313

The fact that the preordering is Archimedean in the compact case was first notedby T. Wörmann [Wö]. An algorithmic proof of Theorem 12.24 was developed by M.Schweighofer [Sw1, Sw2].

The operator-theoretic proof of Theorem 12.36(ii) given above is long knownamong operator theorists; it was used in [Sm6]. The operator-theoretic approach tothe multidimensional moment theory was investigated by F. Vasilescu [Vs1, Vs2].

The representation theorem for Archimedean modules (Theorem 12.35) has along history. It was proved in various versions by M.H. Stone [Stn], R.V. Kadison[Kd], J.-L. Krivine [Kv1], E. Becker and N. Schwartz [BS], M. Putinar [Pu2], andT. Jacobi [Jc]. The version for quadratic modules is due to Jacobi [Jc], while theversion for semirings was proved much earlier by Krivine [Kv1]. A more generalversion and a detailed discussion can be found in [Ms1, Section 5.4]. Lemma 12.34appeared in [BSS]. Putinar [Pu2] has proved that a finitely generated quadraticmodule Q in RdŒ x � is Archimedean if (and only if) there exists a polynomial f 2 Qsuch that the set fx 2 Rd W f .x/ � 0g is compact.

Corollary 12.29 and its non-compact version in Exercise 14.11 below are from[Ls3]. The moment problem with bounded densities is usually called the Markovmoment problem or L-moment problem. In dimension one it goes back to A.A.Markov [Mv1, Mv2], see [AK, Kr2]. An interesting more recent work is [DF]. Themultidimensional case was studied in [Pu1, Pu3, Pu5, Ls3, Ls4].

For compact polyhedra with nonempty interiors Theorem 12.44 was proved by D.Handelman [Hn]. A special case was treated earlier by J.-L. Krivine [Kv2]. A relatedversion can be found in [Cs, Theorem 4]. The general form presented above is takenfrom [PD, Theorem 5.4.6]. Polya’s theorem was proved in [P]. Polya’s original proofis elementary; the elegant proof given in the text is from [Wö]. Proposition 12.54 isa classical result obtained in [HS]. It should be noted that Reznick’s theorem [Re2]is an immediate consequence of the strict Positivstellensatz, see [Sr3, 2.1.8].

Reconstructing the shape of subsets of Rd from its moments with respect to theLebesgue measure is another interesting topic, see e.g. [GHPP] and [GLPR].

Chapter 13The Moment Problem on Closed Semi-AlgebraicSets: Existence

The main subject of this chapter and the next is the moment problem on closed semi-algebraic sets. For a compact semi-algebraic set K.f/ a very satisfactory solutionof the existence problem in terms of the positivity on the preordering T.f/ wasgiven by Theorem 12.25. This result holds for any finite set f of generators whichdefines the semi-algebraic set K.f/. The representing measure is always unique andsupported on K.f/. All these features of the compact case are no longer true fornoncompact sets. In this chapter we are only concerned with existence problems,while determinacy questions are studied in the next chapter.

Let us consider a semi-algebraic set K.f/. Having Theorem 12.25 in mind itis natural to ask when the positivity of a linear functional L on the preorderingT.f/ implies that L is a moment functional. In this case we will say that T.f/has the moment property (MP). If at least one representing measure has supportcontained in K.f/, then T.f/ obeys the strong moment property (SMP). To studywhen these properties hold or fail for a preordering or a quadratic module is themain theme in this chapter. The fundamental result in this respect is the fibretheorem (Theorem 13.10). It is stated and discussed in Sect. 13.3, but the longproof of its main implication is given only in Sect. 13.10. The fibre theorem reducesmoment properties of T.f/ to those for fibre preorderings built by means of boundedpolynomials on the set K.f/. Most of the known general affirmative results on themoment problem for closed semi-algebraic sets can be derived from this theorem.In Sects. 13.4–13.7 we develop a number of applications of the fibre theorem andprovide classes of preorderings satisfying (MP) or (SMP).

On the other hand, one of the new difficulties in dimensions d � 2 is that thepreordering

PRdŒ x �2 does not satisfy (MP), that is, there exist positive functionals

which are not moment functionals. The reason for this is the existence of positivepolynomials in two variables which are not sums of squares of polynomials.Section 13.1 deals with this matter. In Sect. 13.8 the concept of stability is usedto prove the closedness of quadratic modules and to derive classes of quadraticmodules for which (MP) fails. The moment problem on some cubics is studied inSect. 13.9.


315

316 13 The Moment Problem on Closed Semi-Algebraic Sets: Existence

13.1 Positive Polynomials and Sums of Squares

In this section we begin with some simple facts on sums of squares. Then we developMotzkin’s example of a positive polynomial which is not a sum of squares and useit to construct a positive linear functional which is not a moment functional.

We begin with some simple properties of sums of squares of polynomials.

Lemma 13.1 Let p; p1; : : : ; pr 2 RdŒ x � and p ¤ 0. If p DPrjD1 p2j , then

deg. p/ D 2maxfdeg. pj/ W j D 1; : : : ; rg:

Proof Let n denote the maximum of deg. pj/, j D 1; : : : ; r: Clearly, deg. p/ � 2n.We denote by pnj the homogeneous part of degree n of pj. (It may happen that pnj D0 for some j.) Then p2n WDP

j p2nj is the homogeneous part of degree 2n of p. Since

n D max deg. pj/, there is one index, say j D 1, with pn1 ¤ 0. Then p2n.x/ � p2n1.x/on Rd and pn1 ¤ 0. Thus p2n ¤ 0 and hence deg. p/ � 2n. ut

Recall that RdŒ x �m � RŒx1; : : : ; xd�m denotes the vector space of polynomialsp 2 RdŒ x � with deg. p/ � m and

PRŒx�2n is the cone of sums of squares

Pj p2j of

polynomials pj 2 RdŒ x �n. Then, by Lemma 13.1,

RdŒ x �2n \X

RdŒ x �2 D

XRdŒ x �

2n for n 2 N: (13.1)

The vector space RdŒ x �m has dimension d.m/ WD �dCmm

�. A vector space basis of

RdŒ x �m is given by the monomials

x˛ D x˛11 � � � x˛dd ; where ˛ 2 Nm WD f˛ 2 Nd0 W j˛j WD ˛1 C � � � C ˛d � mg:

We order the basis elements of RdŒ x �n in some fixed way and write them as acolumn vector xn. For instance, a possible “natural” ordering is

xn D .1; x1; : : : ; xd; x21; x1x2; : : : ; x2d; x31; : : : ; xn1; : : : ; xnd/T : (13.2)

Proposition 13.2 A polynomial f 2 RdŒ x � is inP

RdŒ x �2n if and only if there existsa positive semidefinite matrix G D .a˛;ˇ/˛;ˇ2Nn with real entries such that

f .x/ D .xn/TGxn �X˛;ˇ2Nn

a˛;ˇx˛Cˇ: (13.3)

Proof First suppose that f D PrjD1 f 2j , where fj 2 RdŒ x �n. We write fj as fj.x/ DP

˛2Nnfj;˛x˛ and define

a˛;ˇ WDrX

jD1fj;˛fj;ˇ and G WD .a˛;ˇ/˛;ˇ2Nn : (13.4)

13.1 Positive Polynomials and Sums of Squares 317

Then

f .x/ DrX

jD1fj.x/

2 DrX

jD1

X˛;ˇ2Nn

fj;˛fj;ˇ x˛Cˇ D

X˛;ˇ2Nn

a˛;ˇx˛Cˇ D .xn/TGxn

which proves (13.3). Obviously, the matrix G is real and symmetric. To prove thatit is positive semidefinite we take a column vector y 2 Rd.n/ and compute

yTGy DX˛;ˇ2Nn

a˛;ˇy˛yˇ DX˛;ˇ2Nn

rXjD1

fj;˛fj;ˇy˛yˇ DrX

jD1

�X˛2Nn

fj;˛y˛

�2� 0:

Conversely, assume that f .x/ D .xn/TGxn, where G is a real positive semidefinitematrix. Let r D rankG. The assertion is trivial for G D 0, so we can assume thatr 2 N. Let D be the diagonal matrix with nonzero, hence positive, eigenvalues�1; : : : ; �r of G and C the matrix with columns u1; : : : ; ur of correspondingorthonormal eigenvectors. Put B WD C

pD . Then B 2 Md.n/;r.R/ and

G DrX

jD1�juj.uj/

T D CDCT D BBT :

Here the first equality holds by the spectral theorem for hermitian matrices, see e.g.(A.12). Setting fj.x/ DP˛2Nn

b˛;jx˛, we have fj 2 RdŒ x �n and

f .x/ D .xn/TBBTxn DX

˛;ˇ2Nn

rXjD1

x˛b˛;jbˇ;jxˇ D

rXjD1

� X˛2Nn

b˛;jx˛

�2D

rXjD1

fj.x/2: ut

Let f 2 PRdŒ x �2n: The matrix G defined by (13.4) is called the Gram matrixassociated with the sum of squares (abbreviated sos) representation f D Pr

jD1 f 2jand formula (13.3) is the correspondingGram matrix representation. Gram matricesare a useful tool for detecting possible sos representations of polynomials, seeExample 16.4. If f .x/ D P

� f�x� 2 P

RŒx�2n, comparing the coefficients of x�

in (13.3) yields

X˛;ˇ2Nn ;˛CˇD�

a˛;ˇ D f� ; where � 2 N2n: (13.5)

The smallest number r appearing in all possible sos representations f D PrjD1 f 2j

of f is called the length of f . The preceding proof shows that the length of f is thesmallest rank of all Gram matrices associated with f , so in particular, it is less thanor equal to d.n/. That is, we have

Corollary 13.3 Each polynomial f 2PRdŒ x �2n is a sum of at most d.n/ D �dCnn

�squares of polynomials of RdŒ x �n.


The assertion of Corollary 13.3 also follows from Carathéodory’s theorem A.35.Obviously,

PRdŒ x �2 is a subset of the cone

Pos.Rd/ D fp 2 RdŒ x � W p.x/ � 0 for x 2 Rdg:

For d D 1 both sets coincides as shown by Proposition 3.1, but for d � 2 we havePRdŒ x �2 ¤ Pos.Rd/. This was already proved in 1888 by D. Hilbert [H1], but the

first explicit example was given only in 1966 by T. Motzkin [Mo]. Another famousexample, the Robinson polynomial, will appear in Section 19.2.

Proposition 13.4 Suppose that 0 < c � 3. Then the polynomial

pc.x1; x2/ WD x21x22.x

21 C x22 � c/C 1 (13.6)

is in Pos.R2/nPRŒx1; x2�2, that is, pc is nonnegative on R2, but it is not a sum ofsquares of polynomials.

Proof From the arithmetic-geometric mean inequality we obtain

x41x22 C x21x

42 C 1 � 3 3

qx41x

22 � x21x42 � 1 D 3x21x22 � cx21x

22;

which in turn implies that pc.x1; x2/ � 0 for all .x1; x2/ 2 R2.Now we prove that pc … PRŒx1; x2�2. Assume to the contrary that pc D P

j q2j ;

where qj 2 RŒx1; x2�. Then we have deg.qj/ � 3 by Lemma 13.1. Since pc.0; x2/ Dpc.x1; 0/ D 1, it follows that the polynomials qj.0; x2/ and qj.x1; 0/ in one variableare constant. Hence each qj is of the form �jC x1x2rj, where �j 2 R and rj 2 RdŒ x �is linear. Comparing the coefficients of x21x

22 in pc DPj q

2j yields

Pj rj.0/

2 D �c.Since c > 0, this is a contradiction. ut

Motzkin’s original example is p3. Since p3.˙1;˙1/ D 0 and p3 � 0 on R2, themimimum of p3 on R2 is zero. From the identity

pc.pcx1;pcx2/ D c3

27p3.p3x1;p3x2/C 1 � c3

27

it follows that the minimum of pc on R2 is 1 � c3

27> 0 if 0 < c < 3.

Using the polynomial p1 we now construct (by some computations) an explicitexample of a positive linear functional on RŒx1; x2� which is not a momentfunctional. The existence of such a functional is obtained later also by separationarguments (see Example 13.52 below).

Let L2 denote the linear functional on RŒx1; x2� defined by

L2.xk1x

l2/ D jm.k=2;l=2/ if k and l are even; L2.x

k1x

l2/ D 0 otherwise;

13.1 Positive Polynomials and Sums of Squares 319

where m is the bijection of N20 onto N and jn are the numbers defined by

m.0; 0/ D 1;m.1; 0/ D 5;m.0; 1/ D 6;m.2; 0/ D 7;m.0; 2/ D 8;m.3; 0/ D 9;m.0; 3/ D 10;m.k; l/ D lC 1C .kC l/.kC lC 1/=2 for .k; l/ 2 N2

0; kC l � 4;j1 D j2 D j3 D 1; j4 D 4; jn D nŠ.nC1/Š for n � 5:

Proposition 13.5 L2 is a positive functional which is not a moment functional.

Proof First we prove that L2. f 2/ > 0 for all f 2 RŒx1; x2�; p ¤ 0. Let us define˛m.k;l/;m.r;s/ WD L2.x

kCr1 xlCs

2 /: If f DPk;l ck;lxk1x

l2, then

L2. f2/ D

Xk;l;r;s

˛m.k;l/;m.r;s/ck;lcr;s : (13.7)

Hence it is enough to show that the quadratic form in (13.7) is positive definite. Forthis it suffices to prove that An WD det .˛k;l/nk;lD1 > 0 for all n 2 N.

We prove by induction that An � 1. We obtain A1 D A2 D A3 D 1 and A4 D 4.Assume that n � 5 and An�1 � 1. A simple computation shows that

max.m.k; l/;m.r; s// > m..kC r/=2; .lC s/=2/ if .k; l/ ¤ .r; s/

and the right-hand side is defined. This implies j˛k;lj � jn�1 for k � n; l � n;.k; l/ ¤ .n; n/ and k; l; n 2 N. Developing the determinant An after the n-th row byusing these facts and the induction hypothesis An�1 � 1 we derive

An � jnAn�1 � .n � 1/.n � 1/Šjnn�1 � jn � nŠjnn�1 C 1 � nŠ.nC1/Š � nŠ.n � 1/ŠnŠ C 1 � 1:

This completes the induction proof. Thus, in particular, L2. f 2/ � 0 for f 2RŒx1; x2�.

A direct verification yields L2. p1/ D �1. Therefore, since p1 � 0 on R2, L2 isnot a moment functional. ut

Clearly, Ld. f / WD L2. f .x1; x2; 0; : : : ; 0//, f 2 RdŒ x �, defines a positive linearfunctional on RdŒ x �, d � 2; which is not a moment functional. An elegant exampleof this kind for d D 2 is sketched in Exercise 13.3.

The preceding examples can be easily used to construct similar examples for thequarter plane and the Stieltjes moment problem in R2. Put

qc.x1; x2/ WD x1x2.x1 C x2 � c/C 1; c 2 .0; 3�: (13.8)


Then pc.x1; x2/ D qc.x21; x22/. Since pc � 0 on R2, it follows that qc � 0 on the

positive quarter plane R2C. But qc does not belong to the preordering

T.x1; x2/ DX

RŒx1; x2�2 C x1

XRŒx1; x2�

2 C x2X

RŒx1; x2�2 C x1x2

XRŒx1; x2�

2:

(If qc were in T.x1; x2/, then, replacing x1 by x21 and x2 by x22, it would follow thatpc 2PRŒx1; x2�2, which is a contradiction.)

Define L02. f / D L2. f .x21; x

22// for f 2 RŒx1; x2�. Then L0

2 is a T.x1; x2/-positivelinear functional on RŒx1; x2�. Since L0.q1/ D L2. p1/ < 0, L0

2 cannot be given by aRadon measure supported on the set K.x1; x2/ D R2C.

13.2 Properties (MP) and (SMP)

In this section, we suppose that A is a finitely generated commutative real unitalalgebra. Recall that in our terminology Radon measures are always nonnegative.

As discussed in Sect. 1.1.2, A is (isomorphic to) the quotient algebra RdŒ x �=Jfor some ideal J of RdŒ x �, the set of characters of A is the real algebraic varietyOA D Z.J /, and OA is a locally compact Hausdorff space. For a quadratic module Q ofA we recall the definition K.Q/ WD fx 2 OA W f .x/ � 0 ; f 2 Qg from (12.10). Further,let MC. OA/ denote the Radon measures � on OA such that each f 2 A is �-integrable.We will use these notions and also the basics of real algebraic geometry developedin Sect. 12.1 without mention in what follows.

Our main concepts are introduced in the following definition.

Definition 13.6 A quadratic module Q of A has the

� moment property (MP) if each Q-positive linear functional L on A is a momentfunctional, that is, there exists a Radon measure � 2MC. OA/ such that

L. f / DZ

OAf .x/ d�.x/ for all f 2 A; (13.9)

� strong moment property (SMP) if each Q-positive linear functional L on A is aK.Q/–moment functional, that is, there is a Radon measure � 2 MC. OA/ suchthat supp� K.Q/ and (13.9) holds.

For d � 2 the preorderingP

RdŒ x �2 does not satisfy (MP); an explicit examplewas given by Proposition 13.5. Obviously, (SMP) implies (MP). That (MP) does notimply (SMP) is shown by the following simple example in dimension d D 1.

Example 13.7 T.x3/ DPRŒx�2 C x3P

RŒx�2 satisfies (MP), but not (SMP).Indeed, since

PRŒx�2 T.x3/, the preordering T.x3/ has (MP) by Hamburger’s

theorem 3.8. It remains to show that (SMP) fails.

13.2 Properties (MP) and (SMP) 321

Let s be a Stieltjes moment sequence which is Hamburger indeterminate (seee.g. the examples in Sect. 4.3). There exists an N-extremal measure� for s such thatsupp� contains a negative number and .�"; "/\ supp� D ; for some " > 0. (Theexistence of such a measure follows easily from Theorem 7.7: There is a unique N-extremal measure which has 0 in its support. Any other N-extremal measure whosesupport contains a negative number has the desired properties.)

We define a measure � 2 MC.R/ by d� D x�2d� and a positive linearfunctional L on RŒx� by L. p/ D R

p d�. Since s is a Stieltjes moment sequence,we have L.x3p2/ D R xp2 d� D Ls.xp2/ � 0 for p 2 RŒx�. Thus, L is T.x3/-positive.

On the other hand, since � is N-extremal, CŒx� is dense in L2.R; �/.Because .1 C x2/d� D .1 C x�2/d� � .1 C "�2/d�, CŒx� is also dense inL2.R; .1 C x2/d�/. Hence the measure � is determinate by Corollary 6.11, that is,� is the only representing measure for L. Since �, hence �, has a negative numberin its support, L has no representing measure supported on RC D K.x3/. Thisshows that T.x3/ does not obey (SMP). Another proof of this fact is given inExample 13.18 below. Note that Stieltjes’ theorem 3.12 implies that the functionalL is not T.x/-positive. ı

The preordering of each compact semi-algebraic set satisfies (SMP) (by The-orem 12.25) and likewise so does each Archimedean quadratic module (by The-orem 12.36). However, deciding whether or not preorderings for noncompactsemi-algebraic sets obey (SMP) or (MP) is much more subtle and the fibre theoremstated below deals with this question.

Suppose that Q is a quadratic module of A. Let Q denote the closure of Q inthe finest locally convex topology of the vector space A, see Appendix A.5 for thedefinition and some properties of this topology. Each linear functional or linearmapping is continuous in this topology. Applying the latter to the multiplicationA�A! A it follows that Q is also a quadratic module and Q is a preordering whenQ is a preordering.

The next lemma gives a simple “dual” characterization of Q .

Lemma 13.8 Q is the set of all f 2 A such that L. f / � 0 for all Q-positive linearfunctionals L on A.

Proof Let QQ denote the set of such elements f 2 A. Suppose f 2 A and f … Q. Then,since Q is closed in the finest locally convex topology, by the separation theorem forconvex sets there is a Q-positive, hence Q-positive, linear functional L on A such thatL. f / < 0. Thus, f … QQ. This proves that QQ Q.

Conversely, let L be a Q-positive linear functional. Since L is continuous in thefinest locally convex topology, L is also Q-positive. Hence L. f / � 0 for f 2 Q, sothat Q QQ. ut

Recall that Qsat D Pos.K.Q// is the saturation of Q. Clearly, Q Pos.K.Q//and Pos.K.Q// is closed in the finest locally convex topology, so we have

Q Q Qsat � Pos.K.Q//:


Haviland’s theorem 1.14 leads to the following reformulations of the properties(SMP) and (MP) in terms of the closure Q of Q in the finest locally convex topology.

Proposition 13.9

(i) (SMP) holds if and only if Q D Qsat � Pos.K.Q//.(ii) (MP) holds if and only if Q D Pos. OA/:Proof We carry out the proof of (i); (ii) is proved by the same reasoning.

Suppose that Q D Pos.K.Q//: Let L be a Q-positive linear functional. Then Lis Q-positive and hence Pos.K.Q//-positive. Thus, by Haviland’s Theorem 1.14, Lcomes from a measure supported on K.Q/: This means that Q obeys (SMP).

Now assume that Q ¤ Pos.K.Q//. Let f0 2 Pos.K.Q//nQ. Then, byLemma 13.8 there is Q-positive linear functional L such that L. f0/ < 0. Sincef0 � 0 on K.Q/ and L. f0/ < 0, L cannot be given by a positive measure supportedon K.Q/. That is, (SMP) does not hold. ut

13.3 The Fibre Theorem

In this section, A is a finitely generated commutative real unital algebra.Let T be a finitely generated preordering of A and let f D f f1; : : : ; fkg be a set of

generators of T. Further, we fix an m-tuple h D .h1; : : : ; hm/ of elements hk 2 A.Let h.K.T// denote the closure of the subset h.K.T// of Rm defined by

h.K.T// D f.h1.x/; : : : ; hm.x// W x 2 K.T/g: (13.10)

For � D .�1; : : : ; �r/ 2 Rm we denote by K.T/� the subset of OA given by

K.T/� D fx 2 K.T/ W h1.x/ D �1; : : : ; hm.x/ D �mg

and by T� the preordering of A generated by the sequence

f.�/ WD f f1; : : : ; fk; h1 � �1; �1 � h1; : : : ; hm � �m; �m � hmg:

Clearly, K.T�/ D K.T/� and K.T/ is the disjoint union of fibre set K.T/�, where� 2 h.K.T//.

Let I� denote the ideal of A generated by h1��1; : : : ; hm��m. Then, by formula(12.9) in Example 12.4,

T� D T C I�

and the preordering T�=I� of the quotient algebra A=I� is generated by

��.f/ WD f��. f1/; : : : ; ��. fk/g;

13.3 The Fibre Theorem 323

where �� W A! A=I� denotes the canonical map.Further, let OI� WD I.Z.I�// denote the ideal of all f 2 A which vanish on the

zero set Z.I�/ of I�. Clearly, I� OI� and Z.I�/ D Z. OI�/. Set

OT� WD T C OI�:

Then OT�= OI� is a preordering of the quotient algebra A= OI�.In general I� ¤ OI� and equality holds if and only if the ideal I is real. The latter

means thatP

j a2j 2 I� for finitely many elements aj 2 A implies that all aj are in

I�. In real algebraic geometry the ideal OI is called the real radical of I.The following fibre theorem is the main result of this chapter. It allows us to

derive (SMP) or (MP) for T from the corresponding properties of fibre preorderings.

Theorem 13.10 Let A be a finitely generated commutative real unital algebra andlet T be a finitely generated preordering of A. Suppose that h1; : : : ; hm are elementsof A that are bounded on the set K.T/. Then the following are equivalent:

(i) T satisfies property (SMP) (resp. (MP)) in A.(ii) T� satisfies (SMP) (resp. (MP)) in A for all � 2 h.K.T//:

(ii)0 OT� satisfies (SMP) (resp. (MP)) in A for all � 2 h.K.T//:(iii) T�=I� satisfies (SMP) (resp. (MP)) in A=I� for all � 2 h.K.T//:

(iii)0 OT�= OI� satisfies (SMP) (resp. (MP)) in A= OI� for all � 2 h.K.T//:

Remark 13.11 The power of Theorem 13.10 can be nicely illustrated by the factthat it contains the main moment problem result (Theorem 12.25) for A D RdŒ x �,T D T.f/; and compact semi-algebraic sets K.T.f// as an immediate consequence.Indeed, since K.T.f// is compact, the coordinate functions xj are bounded onK.T.f//, so they can be taken as functions hj; j D 1; : : : ; d. Then all fibrealgebras A=I�, � 2 h.K.T//; are R and T�=I� D RC obviously has (SMP) inA=I� D R. Hence T D T.f/ obeys (SMP) in RdŒ x � by the implication (iii)!(i) ofTheorem 13.10. ı

To formulate a version of the fibre theorem for quadratic modules let Q be afinitely generated quadratic module of A. We then define the quadratic module

Q� D QC I� and OQ� D QC OI�:

of A and the corresponding fibre set

K.Q/� WD K.Q�/ D fx 2 K.Q/ W h1.x/ D �1; : : : ; hm.x/ D �mg:

Theorem 13.12 Let A be a finitely generated commutative real unital algebra, Q afinitely generated quadratic module of A , and h1; : : : ; hm 2 A. Suppose that there


are ˛j; ˇj 2 R such that ˇj � hj and hj � ˛j are in Q for j D 1; : : : ;m: Set K WDQmjD1Œ˛j; ˇj�. Then the following are equivalent:

(i) Q satisfies property (SMP) (resp. (MP)) in A.(ii) Q� satisfies (SMP) (resp. (MP)) in A for all � 2 K.

(ii)0 OQ� satisfies (SMP) (resp. (MP)) in A for all � 2 K.(iii) Q�=I� satifies (SMP) (resp. (MP)) in A=I� for all � 2 K.

(iii)0 OQ�= OI� satisfies (SMP) (resp. (MP)) in A= OI� for all � 2 K:

Remark 13.13

1. The fibre set K.Q/� may be empty for some � 2 QmjD1Œ˛j; ˇj�. For such � we

have Q� D A and (SMP) holds trivially. However, if � 2 h.K.T//, say � D h.x/with x 2 K.T/, then x 2 K.T/�, so the fibre is not empty.

2. Suppose that ˇj � hj and hj � ˛j are in Q as in Theorem 13.12. Then, sinceevaluations by points x 2 K.Q/ are Q-positive by definition and hence Q-positive, we have ˇj � hj.x/ � 0 and hj.x/ � ˛j � 0. That is, hj is boundedon K.Q/, so the corresponding assumption of Theorem 13.10 is satisfied.

3. In Theorem 13.10 it is only assumed that the polynomials hj are bounded, but notthat ˇj � hj; hj � ˛j are in Q. Therefore, Theorem 13.10 is much stronger thanTheorem 13.12. ıThe implications (i)!(ii), the equivalence (ii)$(ii)0 and the two equivalences

(ii)$(iii) and (ii)0 $(iii)0 of Theorems 13.10 and 13.12 follow immediately fromProposition 13.14 (i),(ii), and (iii), respectively, proved below.

The proofs of the remaining main implication (ii)!(i) of Theorems 13.10and 13.12 are lenghthy and technically involved. They are postponed untilSect. 13.10. Further, a crucial step in the proof of Theorem 13.10 is Propo-sition 12.23, which is based on the Krivine–Stengle Positivstellensatz (Theo-rem 12.3).

Proposition 13.14 Let I be an ideal and Q a quadratic module of A. Let OI be theideal of all f 2 A which vanish on the zero set Z.I/ of I.

(i) If Q satisfies (SMP) (resp. (MP)) in A, so does QC I in A.(ii) QC I satisfies (SMP) (resp. (MP)) in A if and only if QC OI does.

(iii) QCI satisfies (SMP) (resp. (MP)) in A if and only if .QCI/=I does in A=I.

Proof We only carry out the proofs for (SMP). The proofs for (MP) are evensimpler, since no support conditions have to be verified. Let � W RdŒ x � !RdŒ x �=JDA denote the canonical map and QI the ideal QI WD ��1.I/ of RdŒ x �.

(i) Let L be a .QCI/-positive functional on A. Since L is Q-positive and Q obeys(SMP), L is given by a measure� 2MC. OA/ supported on K.Q/. Then QL.�/ WDL.�.�// is a linear functional on RdŒ x �. Since L is I-positive, QL is QI-positiveand hence supp� Z. QI/ by Proposition 12.19. (Note that a functional ispositive on an ideal if and only if it annihilates the ideal.) But Z. QI/ Z.I/,

13.3 The Fibre Theorem 325

so that we have supp� K.Q/\Z.I/ D K.QC I/. That is, QC I satisfies(SMP) in A.

(ii) It suffices to show that both quadratic modules QCI and QC OI have the samenonnegative characters and linear functionals.

For the sets of characters, using the equality Z.I/ D Z. OI/ we obtain

K.QC I/ D K.Q/ \ Z.I/ D K.Q/ \ Z. OI/ D K.QC OI/:

Since QCI QC OI , a .QC OI/-positive functional is trivially .QCI/-positive.Conversely, let L be a .QC I/-positive linear functional on A.We verify that Z. QI/ Z. OI/: Let x 2 Z. QI/. Rd/: Clearly, J QI and

Z. QI/ Z.J /DOA. Let g 2 I and choose Qg 2 QI such that g D �.Qg/. Since xannhilates J and x 2 Z. QI/; we have g.x/ D Qg.x/ D 0: Thus, x 2 Z.I/ DZ. OI/:

Now let f 2 OI. We choose Qf 2 RdŒ x � such that �. Qf / D f . Then f .x/ DQf .x/ for x 2 Z.J /. Hence, since f vanishes on Z. OI/ and Z. QI/ Z. OI/;the polynomial Qf vanishes on Z. QI/. Therefore, by the real Nullstellensatz(Theorem 12.3(iii)), there are m 2 N and g 2 P

RdŒ x �2 such that p WD. Qf /2m C g 2 QI. Upon multiplying p by some even power of Qf we can assumethat 2m D 2k for some k 2 N. Then

�. p/ D f 2k C �.g/ 2 I; where �.g/ 2

XA2:

Being .QC I/-positive, L annihilates I and is nonnegative onP

A2. Hence

0 D L.�. p// D L�f 2

k�C L.�.g//; L.�.g// � 0; L�f 2

k� � 0:This implies that L. f 2

k/ D 0. Since L is nonnegative on

PA2, the Cauchy–

Schwarz inequality holds. By a repeated application of this inequality wederive

jL. f /j2k � L. f 2/2k�1

L.1/2k�1 � L. f 4/2

k�2L.1/2

k�2C2k�1 � : : :� L. f 2

k/L.1/1C��C2k�1 D 0:

Thus L. f / D 0. That is, L annihilates OI. Hence L is .Q C OI/-positive whichcompletes the proof of (ii).

(iii) The assertions are only slight reformulations of Definition 13.6.Let denote the canonical map of A into A=I. Clearly, the character set of

A=I can be identified with Z.I/ D fx 2 OA W f .x/ D 0 for f 2 Ig.Suppose that QCI obeys (SMP) in A. Let QL be a .QCI/=I-positive linear

functional on A=I. Then L WD QLı defines a .QCI/-positive linear functionalon A: Since Q C I has (SMP), the functional L, hence also QL, is given by a


measure of MC. OA/ supported on K.QCI/ D K.Q/\Z.I/ D K..QCI/=I/.This shows that .QC I/=I satisfies (SMP) in A=I.

Conversely, assume that .Q C I/=I has (SMP) in A=I and let L be a.QCI/-positive linear functional on A. Then L is in particular I-positive, so itannihilates I, and there is a well-defined linear functional QL on A=I such thatL D QL ı . Clearly, QL is .QC I/=I-positive. Since .QC I/=I has (SMP), thefunctional QL, and therefore also L, comes from a Radon measure with supportcontained in K..QCI/=I/ D K.QCI/. That is, QCI obeys (SMP) in A. ut

Remark 13.15 The Positivstellensatz for RdŒ x � (Theorem 12.3) is the onlyunproven result from real algebraic geometry we use in this book. In the precedingproof we derived the real Nullstellensatz for A from Theorem 12.3(iii). That the realNullstellensatz holds for the algebra A follows from [PD, Section 4.2]. The aboveproof of assertion (ii) becomes much shorter if we use the latter result. ı

We state the special case Q DP A2 of Proposition 13.14(iii) separately as

Corollary 13.16 If I is an ideal of A, then I CP A2 obeys (MP) (resp. (SMP)) onA if and only if

P.A=I/2 does in A=I.

The following simple fact is used later several times. Of course, it can also bederived directly from Hamburger’s theorem 3.8.

Corollary 13.17 If the algebra A has a single generator, thenP

A2 obeys (MP).

Proof Being single generated, A is isomorphic to a quotient algebra RŒy�=I forsome ideal I of RŒy�. By Hamburger’s theorem 3.8,

PRŒy�2 obeys (MP) in RŒy�

and so does ICPRŒy�2. Therefore, by Corollary 13.16,P.RŒy�=I/2 ŠP A2 has

(MP) in RŒy�=I Š A: ut

13.4 (SMP) for Basic Closed Semi-Algebraic Subsetsof the Real Line

In many applications of Theorem 13.10 the fibres are semi-algebraic subsets of thereal line. In this section, we investigate (SMP) for such sets in detail.

To illustrate the corresponding phenomena we begin with two examples.

Example 13.18 d D 1; f D fx3g;K.f/ D RC.It is obvious that the polynomial x is in Pos.RC/ D Pos.K.f//:We prove that x … T.f/. Assume to the contrary that x D P

j p2j C x3q, where

pj 2 RŒx� and q 2 PRŒx�2. Setting x D 0 yields

Pj pj.0/

2 D 0. Therefore,pj.0/ D 0 and hence pj.x/ D xfj.x/ with fj 2 RŒx� for each j. Inserting this anddividing by x we get 1 D x

Pj f2j C x2q. Setting once more x D 0 we obtain a

contradiction. Thus, x … T.f/ and hence T.f/ ¤ Pos.K.f//.It will be shown by Corollary 13.49 below that the preordering T.f/ is closed

in RŒx� in the finest locally convex topology. Hence T.f/ D T.f/ ¤ Pos.K.f//, sothat T.f/ does not obey (SMP) by Proposition 13.9(i). This was already proved in

13.4 (SMP) for Basic Closed Semi-Algebraic Subsets of the Real Line 327

Example 13.7. There a T.f/-positive linear functional was constructed that cannotbe given by a Radon measure supported on RC.

However, if we replace f D fx3g by Qf D fxg, then K.f/ D K.Qf/ D RC andT.Qf/ D Pos.K.Qf// D Pos.RC/ by formula (3.4) in Proposition 3.2. Thus T.Qf/ has(SMP) by Proposition 13.9(i) (or likewise by Stieltjes’ theorem 3.12).

This shows that, in contrast to the compact case, (SMP) depends in a crucialmanner on the “right” polynomials defining the noncompact semi-algebraic set! ıExample 13.19 d D 1; f D f.1 � x2/3g;K.f/ D Œ�1; 1�.

A similar reasoning as in Example 13.18 (using the zeros˙1 instead) shows thatthe polynomial .1 � x2/ 2 Pos.Œ�1; 1�/ D Pos.K.f// is not in the preordering T.f/.That is, similarly as in Example 13.18 we have T.f/ ¤ Pos.K.f// D Pos.Œ�1; 1�/and also T.Qf/ D Pos.K.Qf// D Pos.Œ�1; 1�/ when we take Qf WD f1 � x2g.

But as K.f/ is compact, T.f/ satisfies (SMP) by Theorem 12.25. Hence we haveT.f/ D Pos.K.f// by Proposition 13.9(i). In particular, T.f/ is not closed in RŒx�. ı

In both examples we have T.f/ ¤ Pos.K.f//, but T.Qf/ D Pos.K.Qf//. To overcomethis difficulty we now define the “right set of generators”.

Let K be a nonempty basic closed semi-algebraic proper subset of R, that is, Kis the union of finitely many closed intervals. (These intervals can be unbounded orpoints.)

Definition 13.20 A finite subset g of RŒx� is called a natural choice of generatorsfor K if g is the smallest set satisfying the following conditions:

� If K contains a least element a (that is, if .�1; a/\K D ;), then .x � a/ 2 g.� If K contains a greatest element a (that is, if .a;1/\K D ;), then .a � x/ 2 g.� If a; b 2 K, a < b, and .a; b/\K D ;, then .x � a/.x � b/ 2 g.

From this definition it is not difficult to see that a choice of natural generatorsalways exists and that it is uniquely determined by the set K. Moreover, we haveK D K.g/. For K D R we set g D f1g.

Let us give some examples for the natural choice of generators:

K D fag [ Œb;C1/, where a < b: g D fx � a; .x � a/.x � b/g;K D Œa; b�[ fcg, where a < b < c: g D fx � a; .x � b/.x � c/; c � xg;K D fag [ fbg, where a < b: g D fx � a; .x � a/.x � b/; b� xg;K D fag: g D fx � a; a � xg:

The next proposition shows that for the natural choice of generators the preorder-ing contains all nonnegative polynomials on K.

Proposition 13.21 Suppose that K is a nonempty basic closed semi-algebraicsubset of R. If g is the natural choice of generators for K, then Pos.K/ D T.g/:

Proof By construction, the natural generators are nonnegative onK, so the inclusionT.g/ Pos.K/ is obvious.

For the converse we prove by induction on the degree of p that p 2 Pos.K/implies p 2 T.g/. If deg. p/ D 0, this is obvious. Assume that it is true if deg. p/ �


n. Suppose that q 2 Pos.K/ and deg.q/ D nC1. If q � 0 on R, then q is inP

RŒx�2

by Proposition 3.1 and so in T.g/. Thus we can assume that q.�/ < 0 for some� 2 R. Since � … K; there are three possible cases.Case 1: K contains a least element a and � < a.

Since q 2 Pos.K/, q has roots in the interval .�; a�. Let c be the least suchroot. Then q D .x � c/p with p 2 Pos.K/ and deg. p/ D n. Since p 2 T.g/ by theinduction hypothesis and .x�a/ 2 g by Definition 13.20, x�c D .x�a/C.a�c/ 2T.g/ and hence q D .x � c/p 2 T.g/:Case 2: K contains a largest element a and � > a.

Let d be the largest root of q in the interval Œa; �/. Then q D .d � x/p withp 2 Pos.K/ and deg. p/ D n, so p 2 T.g/ by the induction hypothesis. ByDefinition 13.20, .a � x/ 2 g. Therefore, d � x D .a � x/ C .d � a/ 2 T.g/,so that q D .d � x/p 2 T.g/.Case 3: There exist a; b 2 K, a < b, such that .a; b/\K D ;:

In this case we take the greatest root d of p in the interval Œa; �/ and the least rootc in the interval .�; b�. Then we can write p D .x� c/.x� d/q with q 2 Pos.K/ anddeg.q/ D n�1; so q 2 T.g/ by the induction hypothesis. We have .x�a/.x�b/ 2 gby Definition 13.20. From Lemma 13.22 below it follows that .x � c/.x � d/ in thepreordering generated by .x�a/.x�b/, so that .x�c/.x�d/ 2 T.g/. Consequently,p D .x � c/.x � d/q 2 T.g/: utLemma 13.22 Suppose that a � b and c; d 2 Œa; b�. There exists a � > 0 such that

.x � c/.x � d/� �.x � a/.x � b/ 2X

RŒx�2:

Proof By a linear transformation we can assume that a D �1 and b D 1. Set� WD sign.cC d/. Since c; d 2 Œ�1; 1� and hence .1 � �c/.1 � �d/ � 0, we obtain

2.x � c/.x � d/� .2 � jcC dj/.x2 � 1/DjcC djx2 � 2.cC d/xC 2C 2cd � jcC djDjcC dj.x2 � 2�xC 1/C 2C 2cd � 2�.cC d/

DjcC dj.x � �/2 C 2.1� �c/.1 � �d/ � 0:

Since c; d 2 Œ�1; 1�, we have � WD 1� jcCdj2� 0 and .x� c/.x� d/� �.x2� 1/ � 0

on R by the preceding inequality. Therefore, .x� c/.x� d/� �.x2 � 1/ 2PRŒx�2

by Proposition 3.1. utThe following elementary fact is used in the proof of Theorem 13.24 below. For

a quadratic f we define w. f / D j�1 � �2j if f has real roots �1; �2 and w. f / D 0 iff has no real roots.

Lemma 13.23 If f1 and f2 are quadratics with positive leading coefficients, then

w. f1 C f2/ � max.w. f1/;w. f2//:

13.4 (SMP) for Basic Closed Semi-Algebraic Subsets of the Real Line 329

Proof Without loss of generality we can assume that w. f2/ � w. f1/ and w. f1/ > 0:Upon translation and scaling it suffices to show the assertion for f1.x/ D x.x � 1/and f2.x/ D c.x � a/.a � .aC b//, where 0 � b � 1, c > 0. Then, for

f1 C f2 D .cC 1/x2 � ..2aC b/cC 1/xC .aC b/ac

we have

w. f1 C f2/ Dp..2aC b/cC 1/2 � 4.cC 1/.aC b/ac

cC 1 :

Since w. f1/ D 1 and w. f2/ D b � 1, we have to prove that w. f1 C f2/ � 1. Aftersquaring and multiplying by .cC 1/2, the assertion w. f1 C f2/ � 1 is equivalent to

..2aC b/cC 1/2 � 4.cC 1/.aC b/ac � .cC 1/2:

A straightforward computation shows that this is equivalent to the inequality

.2aC b � 1/2 C .1 � b2/.cC 1/ � 0:

Since 0 � b � 1, the latter is satisfied, so the assertion w. f1 C f2/ � 1 holds. utThe following theorem characterizes those noncompact semi-algebraic subsets

K.f/ of R for which the preordering T.f/ has (SMP).

Theorem 13.24 Suppose that f D f f1; : : : ; fkg is a finite subset ofRŒx� such that thesemi-algebraic subset K.f/ of R is not compact. The following are equivalent:

(i) T.f/ obeys (SMP).(ii) T.f/ D Pos.K.f//, that is, the preordering T.f/ is saturated.

(iii) f contains positive multiples of all polynomials of the natural choice ofgenerators g of K.f/.

Proof By Proposition 13.9(i), (SMP) holds if and only if T.f/ D Pos.K/. Since K.f/is not compact and semi-algebraic, it contains an unbounded interval. Therefore,T.f/ is closed by Proposition 13.51 proved in Sect. 13.8. We take Proposition 13.51for granted in this proof. Then (i)$(ii).

(iii)!(ii) (iii) implies that T.g/ T.f/. Obviously, T.f/ Pos.K.f//. Propo-sition 13.21 yields Pos.K.f// D T.g/: Putting these facts together we obtainT.f/ D Pos.K.f//:

(ii)!(iii) As already noted, K.f/ contains an unbounded interval. Upon replacingx by �x we can assume that Œc;C1/ K.f/ for some c 2 R. Also we can assumethat all elements of f are not constant. Each f 2 T.f/ is of the form

f DX

ef e11 � � � f ekk �e with �e 2

XRdŒ x �

2: (13.11)


The summation in (13.11) is over e D .e1 : : : ; ek/, where e1; : : : ; ek 2 f0; 1g:All fj are nonnegative on Œc;C1/, so they have positive leading coefficients.

Hence the degree of f in (13.11) is equal to the maximum of the degrees of thesummands.

Case 1: K.f/ contains a least element a.Then f WD x� a 2 Pos.K.f// D T.f/. Since f is linear, all nonzero summands in

the representation (13.11) of f are multiples of linear fj by positive constants. Sincea 2 K.f/, fj.a/ � 0 for each such fj. Since f .a/ D 0, at least one such fj vanishes ata. Hence fj.x/ D �.x � a/ with � > 0, that is, f contains a positive multiple of thenatural generator x � a.

Case 2: There are a; b 2 K.f/, a < b, such that .a; b/\K.f/ D ;:Then f WD .x � a/.x � b/ 2 Pos.K.f// D T.f/ and f .x/ < 0 on .a; b/. Hence

the degrees of all summands in the representation (13.11) of f do not exceed 2.We omit all summands which are nonnegative on .a; b/: Each linear fj is increasing(because fj � 0 on Œc;C1/) and satisfies fj.a/ � 0 (since a 2 K.f/). Hence linearfj and their products are positive on .a; b/. Since f .x/ < 0 on .a; b/, there existquadratic polynomials f1; : : : fr in the set f and positive numbers ˛1; : : : ; ˛r such thatf .x/ � ˛1f1.x/C � � �˛rfr.x/ on .a; b/ and each such fj has at least one negative valueon .a; b/. Since fj � 0 on Œc;C1/ K.f/ and at a; b 2 K.f/, the polynomial fj,hence ˛jfj, has two real zeros in Œa; b�, that is, w.˛jfj/ � b � a. From Lemma 13.23it follows that w. f / D b � a is at most the maximum of w.˛jfj/, j D 1; : : : ; r. Thusw.˛jfj/ D b � a for at least one j. Hence ˛jfj D �.x � a/.x � b/ for some � > 0, soa positive multiple of the natural generator .x � a/.x � b/ belongs to f. ut

13.5 Application of the Fibre Theorem: Cylinder Setswith Compact Base

Perhaps the most natural application of the fibre theorem concerns subsets ofcylinders with compact base.

Proposition 13.25 Let C be a compact set in Rd�1, d � 2; and let f be a finitesubset of RdŒ x �. Suppose that the semi-algebraic subset K.f/ of Rd is contained inthe cylinder C �R. Then the preordering T.f/ has (MP). If C is a semi-algebraic setin Rd�1 and K.f/ D C �R, then T.f/ satisfies (SMP).

Proof Define hj.x/ D xj for j D 1; : : : ; d�1. Since K.f/ C�R and C is compact,the polynomials hj are bounded on K, so the assumptions of Theorem 13.10 arefulfilled. Then all fibres K.f/� are subsets of .�1; : : : ; �d�1/ � R, the preorderingT.f/� contains

PRŒxd�2, and the quotient algebra RdŒ x �=I� is an algebra of

polynomials in the single variable xd. Hence, by Corollary 13.17,P.RdŒ x �=I�/2

obeys (MP) in RdŒ x �=I� and so does T=I�. Therefore, T.f/ has (MP) by theimplication (ii)!(i), and likewise by (iii)!(i), of Theorem 13.10.

13.5 Application of the Fibre Theorem: Cylinder Sets with Compact Base 331

If K.f/ D C �R, the fibres for � 2 h.K.f// are equal to .�1; : : : ; �d�1/ �R andT.f/� DPRŒxd�2. Hence the T.f/� satisfy (SMP) and so does T.f/. ut

We restate this result in the special case of a strip Œa; b� �R in R2.

Example 13.26 Let a; b 2 R, a < b, d D 2, and f D f.x1 � a/.b � x1/g. Then

K.f/ D Œa; b� �R ; T.f/ DX

RŒx1; x2�2 C .x1 � a/.b� x1/

XRŒx1; x2�

2:

By Proposition 13.25, T.f/ obeys (SMP). That is, given a linear functional L onRŒx1; x2�, there exists a Radon measure � on R2 supported on Œa; b� �R such thatp is �-integrable and

L. p/ DZ b

a

ZR

p.x1; x2/ d�.x1; x2/ for all p 2 RŒx1; x2�

if and only if

L.q21 C .x1 � a/.b � x2/q22/ � 0 for all q1; q2 2 RŒx1; x2�: ı

Let us return to Proposition 13.25 and assume that the semi-algebraic subsetK.f/ of Rd is only a proper subset of C � R. Then T.f/ does not satisfy (SMP) ingeneral. Recall that, by Theorem 13.10, T.f/ obeys (SMP) if (and only if) all fibrepreorderings T.f/�, or equivalently, all preorderings T�=I� of the quotient algebrasRdŒ x �=I� do. If a fibre set K.f/� is compact, we know that T.f/� D T.f.�// has(SMP) by Theorem 12.25. Now let us look at the case when a fibre set K.f/� isnot compact. If we take the polynomials hj D xj, j D 1; : : : ; d � 1, as in theproof of Proposition 13.25, the quotient algebra RdŒ x �=I� is (isomorphic to) thepolynomial algebra RŒxd�. Therefore, by Theorem 13.24, the preordering T�=I� inRŒxd� has (SMP) if and only if the set ��.f/ contains positive constant multiples ofall natural choice generators for the corresponding semi-algebraic subset K.T�=I�/of R. That is, in order to conclude (SMP) for T.f/ all noncompact fibres require acareful inspection of the sequence ��.f/.

We illustrate the preceding discussion with four examples. All sets are containedin the strip Œ0; 1� �R, so that (MP) is always satisfied by Proposition 13.25.

Example 13.27 f1.x/ D x1; f2.x/ D 1 � x1; f3.x/ D x32 � x22 � x1; f4.x/ D 4 � x1x2.Then h1.x/ D x1 is bounded and h1.K.f// D Œ0; 1�. The fibres for � 2 .0; 1� are

compact, so the preordering T� has (SMP). The fibre set at � D 0 is f0g [ Œ1;C1/and the sequence �0.f/ is f0; 1; x32� x22; 4g. Since �0.f/ does not contain multiples ofall natural choice generators for f0g [ Œ1;C1/, T.f/ does not have (SMP). ıExample 13.28 f1.x/ D x1; f2.x/ D 1 � x1; f3.x/ D 1 � x1x2; f4.x/ D x32.

Taking again h1.x/ D x1, we have h1.K.f// D Œ0; 1�. All fibres at � 2 .0; 1�are compact, so they obey (SMP). The fibre set at � D 0 is Œ0;C1/ and thecorresponding sequence �0.f/ D f0; 1; 0; x32g does not contain a multiple of


the natural choice generator x2 for Œ0;C1/. Hence T.f/ does not satisfy (SMP).However, if we replace f4 by Qf4.x/ D x2, then T.f/ has (SMP). ıExample 13.29 f1.x/ D x1; f2.x/ D 1 � x1; f3.x/ D x1x2 � 1; f4.x/ D 2 � x1x2.

The set K.f/ is not compact and it is the part of the strip between the twohyperbolas x1x2 D 1 and x1x2 D 2. Then h1.x/ D x1 and h2.x/ D x1x2 are boundedon K.f/ and h.K.f// D .0; 1� � Œ0; 1� is not closed. All fibre sets are points. Sincethey are compact, all fibre preorderings obey (SMP). Therefore, T.f/ has (SMP). ıExample 13.30 f1.x/ D x1; f2.x/ D 1 � x1; f3.x/ D 1 � x1x2:

Then K.f/ is the part of the strip below the hyperbola x1x2 D 1. Set h1.x/ D x1.Then h1.K.f// D Œ0; 1�. The fibre set at � D 0 is the whole x2-axis, so T0 has(SMP). For � 2 .0; 1�, the fibre set is .�1; ��1� and a multiple of its natural choicegenerator ��1 � x2 belongs to ��.f/ D f0; 1; 1� �x2g, so T� also has (SMP). HenceT.f/ satisfies (SMP). ı

The next proposition is also about cylinder sets with compact base.

Proposition 13.31 Suppose that A is a finitely generated real unital algebra withcompact character space OA. Let B be the tensor product of A and the polynomialalgebra RŒx� in a single variable x. Then the preorderings

PA2 and

PB2 obey

(SMP) in A and B, respectively.

Proof Let h1; : : : ; hm be a set of generators of A and consider a nonempty fibre for Aand B, respectively. Then the corresponding fibre sets are points resp. a real line. Thefibre algebra A=I� is R and the fibre algebra B=I� is RŒx�. In both cases

P.A=I�/2

andP.B=I�/2 have (SMP) and so have

PA2 and

PB2 in A and B, respectively, by

Theorem 13.10 (iii)!(i). ut

13.6 Application of the Fibre Theorem: The RationalMoment Problem on Rd

Let us begin with some notation and preliminaries. For a subset D RdŒ x � we put

ZD WD [q2D Z.q/; where Z.q/ D fx 2 Rd W q.x/ D 0g:

Let D.RdŒ x �/ denote the family of all multiplicative subsets D of RdŒ x � (that is,f1; f2 2 D implies f1f2 2 D) such that 1 2 D and 0 … D. Fix D 2 D.RdŒ x �/. Then

AD WD D�1RdŒ x �

is a real unital algebra of rational functions which contains RdŒ x � as a subalgebra.Obviously, if D is finitely generated, so is the algebra AD. The elements of AD arewell-defined real-valued functions on RdnZD . The following lemma shows that thecharacters of AD are just the point evaluations on this set.

13.6 Application of the Fibre Theorem: The Rational Moment Problem on Rd 333

Lemma 13.32 Define �t. f / D f .t/ for t 2 RdnZD and f 2 AD . The character setof the algebra AD is cAD D f�t W t 2 RdnZDg.Proof Clearly, �t is a well-defined character on AD for t 2 RdnZD , so �t 2 cAD.

Conversely, suppose that � 2 cAD. Put tj D �.xj/ for j D 1; : : : ; d. Then we havet D .t1; : : : ; td/ 2 Rd and �. p.x// D p.�.x1/; : : : �.xd// D p.t/ for p 2 RdŒ x �:For q 2 D we obtain 1 D �.1/ D �.qq�1/ D �.q/�.q�1/ D q.t/�.q�1/: Henceq.t/ ¤ 0 for all q 2 D, that is, t 2 RdnZD . Further, �.q�1/ D q.t/�1. Therefore wederive �. pq / D �. p/�.q�1/ D p.t/q.t/�1 D �t. pq /. Thus � D �t. ut

Now let T be a preordering of AD and h D fh1; : : : ; hmg an m-tuple of elementshj 2 AD . For � 2 Rd let I� be the ideal of AD generated by hj � �j, j D 1; : : : ;m:Recall that the subset h.K.T// of Rm was defined by (13.10).

We consider the following assumptions:

(i) The functions h1; : : : ; hm 2 A are bounded on the set K.T/.(ii) For each � 2 h.K.T// there are a finitely generated set E� 2 D.RŒy�/ and a

surjective algebra homomorphism

� W E�1� RŒy�! D�1RdŒ x �=I� � AD=I�:

Our main result of this section is the following existence theorem for themultidimensional rational moment problem.

Theorem 13.33 Suppose thatD 2 D.RdŒ x �/ is finitely generated and T is a finitelygenerated preordering of the algebra AD D D�1RdŒ x �: Assume (i) and (ii).

Then T satisfies (MP), that is, for each T-positive linear functional L on AD thereis a Radon measure � on cAD Š RdnZD such that f is �-integrable and

L. f / DZbAD

f .x/ d�.x/ for f 2 A:

The proof of this theorem is based on the following result which deals with theone-dimensional case. Let RŒy� denote the real polynomials in a single variable y.

Proposition 13.34 Suppose that E 2 D.RŒy�/ is finitely generated. Let B be thereal unital algebra B D E�1RŒy�: Then the preordering

PB2 satisfies (MP), that

is, for each positive linear functional L on B there exists a Radon measure � on thelocally compact Hausdorff space Y WD RnZE such that B L1.Y; �/ and

L. f / DZYf .y/ d�.y/ for f 2 B: (13.12)

Proof Since E is finitely generated, ZE is a finite set, say ZE D fy1; : : : ; ykg. HenceY WD RnZE is a locally compact Hausdorff space in the induced topology from R.Since q.y/ ¤ 0 for q 2 E and y 2 Y , B is a linear subspace of C.YIR/:


We show that B is an adapted space. This means that we have to check the threeconditions (i)–(iii) of Definition 1.5. Since the unital algebra B is the span of itssquares, we have B D BC � BC, so condition (i) is satisfied. (ii) holds with f WD 1.We verify (iii). We choose qj 2 E such that qj.yj/ D 0. Then h WD q21 : : : q

2k 2 E and

h.y/ D 0 for y 2 ZE . If ZE is empty, we set h.y/ D 1C y2. Now let f D pq 2 BC,

where p 2 RŒy� and q 2 E . Further, let " > 0 be given. Then g WD fh 2 BC: Since hvanishes on ZE and limjyj!C1 h.y/ D C1, there exists a compact subset K" of Ysuch that f .y/ � "g.y/ on YnK". This proves (iii). Thus, B is an adapted space.

Next we prove that BC PB2. Let f 2 BC, that is, f .y/ � 0 for y 2 Y . We writef as f D p

q 2 B, where q 2 E and p 2 RŒy�. Then q2f D pq � 0 on Y D RnZEand hence on R, since ZE is empty or finite. The nonnegative polynomial pq in one(!) variable is a sum of squares in RŒy� by Proposition 3.1, so that pq DPj p

2j with

pj 2 RŒy�. Therefore, since q 2 E , we get f DPj .pjq /2 2PB2:

Let f 2 BC. Then f 2 P B2 and hence L. f / � 0, because the functional L ispositive by assumption. Thus L is a BC-positive linear functional on the adaptedsubspace B of C.YIR/. Hence Theorem 1.8 applies and yields the assertion. utProof of Theorem 13.33 Let us fix � 2 h.K.T// and abbreviate B� WD E�1

� RŒy�:By Proposition 13.34,

P.B�/

2 has (MP) in the algebra B�. We denote by J � thekernel of the homomorphism � W B� ! AD=I�. Since

P.B�/

2 has (MP) in B�, it isobvious that J � CP.B�/2 has (MP) in B�. Therefore

P.B�=J �/2 satisfies (MP)

in B�=J � by Corollary 13.16. Since the algebra homomorphism � is surjective, thealgebra B�=J � is isomorphic to AD=I�. Hence

P.AD=I�/2 has (MP) in AD=I� as

well. Consequently, sinceP.AD=I�/2 T=I�, the preordering T=I� obeys (MP)

in AD=I�. Therefore, T has (MP) in AD by Theorem 13.10 (iii)!(i). utThe general fibre theorem fits nicely to the multidimensional rational moment

problem, because in general algebras of rational functions contain more boundedfunctions on K.T/ than polynomial algebras.

The use of Theorem 13.33 is illustrated by some examples. The ideas developedtherein can be combined to treat more involved examples. Throughout we supposethat the corresponding sets D are finitely generated.

Example 13.35 First let d D 2. Suppose that D 2 D.RŒx1; x2�/ contains x1 � ˛and that the semi-algebraic set K.T/ is a subset of f.x1; x2/ W jx1 � ˛j � cg forsome ˛ 2 R and c > 0. Then h1 WD .x1 � ˛/�1 2 AD is bounded on K.T/, soassumption (i) holds. Let � 2 h1.K.T//. Then we have x1 D ��1C ˛ in the algebra.AD/�, so .AD1/� consists of rational functions in x2 with denominators from somefinitely generated set E� 2 D.RŒx2�/. Hence assumption (ii) is also satisfied. ThusTheorem 13.33 applies to AD.

The above setup extends at once to d 2 N; d � 2; if we assume that xj�˛j 2 Dand jxj�˛jj � c on K.T/ for some ˛j 2 R, c > 0, and j D 1; : : : ; d � 1: ıExample 13.36 Suppose D 2 D.RdŒ x �/ is generated by the polynomials qj D 1Cx2j ; j D 1; : : : ; d. Let T D P

.AD/2. Then, by Lemma 13.32, we have K.T/ D

13.7 Application of the Fibre Theorem: A Characterization of Moment. . . 335

cAD D Rd. Setting hj D qj.x/�1, hdCj D xjqj.x/�1 for j D 1; : : : ; d, all hl 2 AD arebounded on K.T/.

Let � 2 h.K.T//: Then �j D qj.�/�1 ¤ 0 and �dCj D �jqj.�/�1; hence we havexj D �dCj�

�1j , j D 1; : : : ; d, in the fibre algebra .AD/�: Thus, .AD/� D R: Taking

E� D f1g, we have E�1� RdŒ x � D RdŒ x �, so (i) and (ii) are obviously satisfied.

Therefore,P.AD/2 obeys (MP) by Theorem 13.33. That is, each positive linear

functional on the algebra AD is given by some positive measure on cAD D Rd.The same conclusion and almost the same reasoning remain valid if we assume

instead that D is generated by the single polynomial q D 1C x21 C � � � C x2d: In thiscase we set hj D xjq.x/�1 for j D 1; : : : ; d and hdC1 D q.x/�1: ıExample 13.37 Suppose D 2 D.RdŒ x �/ contains the polynomials qj D 1C x2j forj D 1; : : : ; d � 1. Let T D P

.AD/2. Then hj WD qj.x/�1 and hdCj�1 WD xjqj.x/�1for j D 1; : : : ; d � 1 are in AD and bounded on K.T/ D cAD. Arguing as inExample 13.36 we conclude that we have xj D �dCj�1��1

j for j D 1; : : : ; d�1 in thealgebra .AD/�: Therefore the fibre algebra .AD/� is an algebra E�1

� RŒxd� of rationalfunctions in the single variable xd for some finitely generated set E� 2 D.RŒxd�/.

Then, again by Theorem 13.33,P.AD/2 satisfies (MP). The same is true if we

assume instead that the polynomial q D 1C x21 C � � � C x2d�1 is in D 2 D.RdŒ x �/. ı

13.7 Application of the Fibre Theorem: A Characterizationof Moment Functionals

In this section, we derive a theorem which characterizes moment functionals on Rd

in terms of extensions. It will be used in the proof of Theorem 15.14 below, but it isalso of interest in itself. Throughout we assume that d � 2.

Let A denote the real algebra of functions on .Rd/� WD Rdnf0g generated by thepolynomial algebra RdŒ x � and the functions

fkl.x/ WD xkxl.x21 C � � � C x2d/

�1; k; l D 1; : : : ; d: (13.13)

Clearly, these functions satisfy the relations

f11 C f22 C � � � C fdd D 1; (13.14)

fijfkl D fikfjl for i; j; k; l D 1; : : : ; d: (13.15)

Thus, the algebra A has d C �dC12

�generators x1; : : : ; xd; f11; f12; : : : ; fdd and the

elements of A are precisely all functions of the form

g. p.x/; f11.x/; : : : ; fdd.x//; (13.16)


where p 2 RdŒ x � and g is a real polynomial in 1C �dC12

�variables. Of course, given

an element of A the polynomial g is not uniquely determined.We denote by Sd�1 the unit sphere of Rd and by Sd�1C the set of all t D .t1; : : : ; td/

of Sd�1 for which the first nonzero coordinate tj is positive.The next lemma describes the character set OA of A.

Lemma 13.38

(i) For x 2 .Rd/� the point evaluation of functions at x is a character �x ofA such that x D .�x.x1/; : : : ; �x.xd// ¤ 0. Each character of A satisfying.�.x1/; : : : ; �.xd// ¤ 0 is of this form.

(ii) For t 2 Sd�1 there exists a unique character �t of A such that

�t.xj/ D 0 and �t. fkl/ D fkl.t/ for j; k; l D 1; : : : ; d: (13.17)

Each character � of A for which �.xj/ D 0 for all j D 1; : : : ; d is of the form�t with uniquely determined t 2 Sd�1C .

(iii) The set OA is the disjoint union of the set f�x W x 2 .Rd/�g and f�t W t 2 Sd�1C g.Proof

(i) The first assertion is obvious.We prove the second assertion. For this let � be a character of A such that

x WD .�.x1/; : : : ; �.xd// ¤ 0. The identity .x21C� � �Cx2d/fkl D xkxl implies that

.�.x1/2 C � � � C �.xd/2/�. fkl/ D �.xk/�.xl/

and therefore

�. fkl/ D .�.x1/2 C � � � C �.xd/2/�1�.xk/�.xl/ D fkl.x/:

Thus � acts on the generators xj and fkl, hence on the whole algebra A, by pointevaluation at x, that is, we have � D �x.

(ii) Fix t 2 Sd�1: Since lim"!C0 fkl."t/ D fkl.t/ and lim"!C0 p."t/ D p.0/ forp 2 RdŒ x �, it follows that the limit

�t.g/ WD lim"!C0 g."t/ D lim

"!C0 �"t.g/

exists for all g 2 A. Since �"t is a character on A, so is �t. By construction wehave �t. fkl/ D fkl.t/ and �t.xj/ D 0 for all j; k; l D 1; : : : ; d.

Conversely, let � be a character of A such that �.xj/ D 0 for j D 1; : : : ; d.Set �kl WD �. fkl/. Since � is a character, (13.14) and (13.15) imply that

�11 C �22 C � � � C �dd D 1; (13.18)

�ij�kl D �ik�jl for i; j; k; l D 1; : : : ; d: (13.19)

13.7 Application of the Fibre Theorem: A Characterization of Moment. . . 337

First we note that �ii � 0 for all i. Assume to the contrary that �ii < 0 forsome i. From (13.18) there exists a j such that �jj > 0. Then �2ij D �ii�jj < 0

by (13.19), which is a contradiction.Set ti D

p�ii, i D 1; : : : ; d. Then t D .t1; : : : ; td/ 2 Sd�1 by (13.18). From

(13.19) we then obtain that �2ij D t2i t2j for all i; j. Therefore, if ti D 0 for some

i, then �ij D 0 and �t. fij/ D �ij D 0 D titj for all j, and this remains validif tj is changed. By leaving out all indices with ti D 0 we can assume withoutloss of generality that all ti are nonzero. Further, since �2ij D t2i t

2j ; there exists

an "ij 2 f�1; 1g such that �ij D "ijtitj for i ¤ j; i; j D 1; : : : ; d. If "1;j D �1 forsome j D 2; : : : ; d, we replace tj by �tj. Then �1j D t1tj for all j D 2; : : : ; d.Further, using (13.19) we obtain

�ij�1j D "ijtitjt1tj D �1i�jj D t1tit2j :

Since all tk are nonzero, "ij D 1. Thus �. fij/ D �ij D titj, that is, �. fij/ D�t. fij/ for all i; j. Since � and �t coincide on generators of A, we have � D �t

on A. By construction, t1 > 0, so that t 2 Sd�1C .We verify the uniqueness assertion. For let �t D �Qt with t; Qt 2 Sd�1C . Since

t2i D �t. fii/ D �Qt. fii/ D Qt2i ; the first nonvanishing indices for t and Qt are thesame, say j. Then tj D Qtj > 0, since t; Qt 2 Sd�1C . Further, tktj D �t. fkj/ D�Qt. fkj/ D QtkQtj D Qtktj: Therefore, since tj ¤ 0, we get Qtk D tk, so that t D Qt.

(iii) follows at once by combining (i) and (ii). utTheorem 13.39 The preordering

PA2 of the algebra A has (MP), that is, for each

positive linear functional L on A there exist Radon measures �0 2 MC.Sd�1/ and�1 2MC.Rd/ such that �1.f0g/ D 0 and for all g 2 A of the form (13.16) we have

L.g. p.x/; f11.x/; : : : ; fdd.x/// D (13.20)ZSd�1

g. p.0/; f11.t/; : : : ; fdd.t// d�0.t/CZRdnf0g

g. p.x/; f11.x/; : : : ; fdd.x// d�1.x/:

Proof It suffices to prove thatP

A2 obeys (MP). The other assertions follow fromthe definition of (MP) and the form of the character set given in Lemma 13.38.

From the description of OA it is obvious that the functions fkl, k; l D 1; : : : ; d; arebounded on OA, so we can take them as functions hj in Theorem 13.10. Let us fix anonempty fibre for � D .�kl/, where �kl 2 R for all k; l. In the quotient algebraA=I� of A by the fibre ideal I� we have �. fkl/ D �kl for all � 2 OA.

First let � D �x, where x 2 .Rd/�. Then �x. fkl/ D fkl.x/ D �kl. By (13.14)we have 1 D P

k fkk.x/ DP

k �kk, so there exists a k such that �kk ¤ 0. From theequality �kk D fkk.x/ D x2k.x

21C� � �Cx2d/�1 we obtain xk ¤ 0. Thus �kl

�kkD fkl.x/

fkk.x/D xl

xk,

so that

xl D �kl��1kk xk for l D 1; : : : ; d: (13.21)


If � D �t for t 2 Sd�1, then �.xl/ D �.xk/ D 0, so (13.21) holds trivially. That is, inthe algebra A=I� we have the relations (13.21) and fkl D �kl. This implies that thequotient algebra A=I� is generated by the single polynomial xk. Hence

P.A=I�/2

satisfies (MP) in A=I� by Corollary 13.17. Therefore, by Theorem 13.10 (iii)!(i),the preordering T DP A2 obeys (MP) in A. ut

Now we are ready to prove the main result of this section.

Theorem 13.40 A linear functional L on RdŒ x � is a moment functional if and onlyif it has an extension to a positive linear functional L on the larger algebra A.

Proof Assume first that L has an extension to a positive linear functional L on A.By Theorem 13.39, L has the form described by Eq. (13.20). We define a Radonmeasure � on Rd by

�.f0g/ D �0.Sd�1/; �.Mnf0g/ D �1.Mnf0g/:

Then � 2MC.Rd/, since �1 2MC.Rd/. Let p 2 RdŒ x �. Setting g.1; 0; : : : ; 0/ D1, we have g. p; f11; : : : ; fdd/ D p and it follows from (13.20) that

L. p/ D QL. p/ D �0.f0g/p.0/CZRdnf0g

p.x/ d�1.x/ DZRd

p.x/ d�.x/:

That is, L is a moment functional on RdŒ x � with representing measure �.Conversely, suppose that L is a moment functional on RdŒ x � and let � be a

representing measure. Fix a point t 2 Sd�1. By Lemma 13.38(ii), �t is a characterof A and �t. f / D f .0/ for f 2 RdŒ x �. Therefore, for f 2 RdŒ x �, we obtain

L. f / D �.f0g/�t. f /CZRdnf0g

f .x/ d�.x/: (13.22)

For f 2 A we define L. f / by the right-hand side of (13.22). Since the functions fklare bounded on Rdnf0g, the integral in (13.22) exists for all f 2 A and it is a positivefunctional on A. The character �t is obviously a positive functional on A. Hence Lis a positive linear functional on A which extends L by (13.22). ut

13.8 Closedness and Stability of Quadratic Modules

While the preceding sections dealt with affirmative results for (SMP) or (MP), theaim of this section is to develop examples where (MP) fails and to provide sometools for proving this. By Proposition 13.9, a quadratic moduleQ of RdŒ x � has (MP)if and only if Q D Pos.Rd/. Thus, proving that (MP) fails requires a polynomial inPos.Rd/ that does not belong to the closure (!) of Q. This leads to the problem ofwhen a quadratic module is closed in the finest locally convex topology. Let us lookat the simplest example

PRdŒ x �2. The closedness of

PRdŒ x �2 can be derived

13.8 Closedness and Stability of Quadratic Modules 339

from the closedness ofP

RdŒ x �2n in the finite-dimensional space RdŒ x �2n combinedwith the simple, but crucial observation RdŒ x �2n \ PRdŒ x �2 D P

RdŒ x �2n.An elaboration of the latter relation leads to the notion of stability for quadraticmodules.

In this section, A is a finitely generated commutative real unital algebra.We equip the countable-dimensional vector space A with the finest locally convextopology (see Appendix A.5). Closedness of sets in A always refers to this topology.On each finite-dimensional subspace it inherits the unique norm topology.

Let W be a subspace of A. We denote byP

W2 the set of sums of squares ofelements of W. For g1; : : : ; gk 2 A, we put

X.WI g1; : : : ; gk/ WD

XW2 C g1

XW2 C � � � C gk

XW2: (13.23)

If n D dimW <1, thenP

W2 andP.WI g1; : : : ; gk/ are contained in subspaces

of dimensions at most .nC1/n2

and .nC1/n.kC1/2

, respectively. (Indeed, if a1 : : : ; an is abasis of W, each square a2 of a 2 W is in the span of elements akal, k � l.)

Definition 13.41 A quadratic module Q of A generated by g1; : : : ; gk 2 A is calledstable if for each finite-dimensional subspace V of A there exists a finite-dimensionalsubspace WV of A such that

V \ Q X

.WV I g1; : : : ; gk/: (13.24)

Lemma 13.42 Definition 13.41 is independent of the choice of generators of Q.

Proof By induction it suffices to show that the condition in Definition 13.41 ispreserved if one element g 2 Q is added to the generators g1; : : : ; gk of Q. If thecondition holds for g1; : : : ; gk, it trivially holds for g1; : : : ; gk; g.

Conversely, suppose that it holds for g1; : : : ; gk; g. Let V be a given finite-dimensional subspace of A. Then there is a finite-dimensional subspace W0 such thatV \ Q P

.W0I g1; : : : ; gk; g/: Since g 2 Q and Q is generated by g1; : : : ; gk, wehave g D �0 C g1�1 C : : : gk�k with �j 2 P A2. We choose a finite-dimensionalsubspace W1 such that 1 2 W1 and all �j are in

PW21 . If WV is the finite-

dimensional subspace spanned by w0w1, where w0 2 W0 and w1 2 W1, thenW0 WV and

gw20 D �0w20 C g1�1w20 C : : : gk�kw20 2

X.WV I g1; : : : ; gk/

for w0 2 W0. Hence V \Q P.WV I g1; : : : ; gk/. utFor the next results we assume that .An/n2N is a sequence of linear subspaces of

A such that

A D1[nD1

An; where An AnC1; dimAn <1 for n 2 N: (13.25)


Our standard example is An D RdŒ x �n � fp 2 RdŒ x � W deg. p/ � ng for A D RdŒ x �.The main motivation for the concept of stability stems from the following fact.

Proposition 13.43 Let Q be a finitely generated quadratic module of A withgenerators g1; : : : ; gk. Let .An/n2N be a sequence of subspaces of A satisfying(13.25). If Q is stable and the set

P.AnI g1; : : : ; gk/ is closed (in the norm topology

of some finite-dimensional subspace of A which containsP.AnI g1; : : : ; gk/) for

each n 2 N, then Q itself is closed in the finest locally convex topology of A.

Proof Since Q is stable, for each An there is a finite-dimensional subspace WAn suchthat An \ Q P

.WAn I g1; : : : ; gk/. By (13.25) there is an mn 2 N such that Amn

contains a vector space basis of WAn . Hence WAn Amn . Then

An \Q .WAn I g1; : : : ; gk/ X

.Amn I g1; : : : ; gk/ Q

and therefore An \ Q D An \ P.Amn I g1; : : : ; gk/. SinceP.Amn I g1; : : : ; gk/ is

closed by assumption, so is An \P.Amn I g1; : : : ; gk/ D An \ Q in An. Hence, byProposition A.28, Q is closed in the finest locally convex topology of A. ut

Often it is convenient to rephrase the stability of quadratic modules in termsof the decomposition (13.25) of A. A quadratic module Q of A with generatorsg1; : : : ; gk is stable if and only if the following condition holds:

(�) There exists a map l W N0 ! N0 such that, for each n 2 N0 and p 2 Q\ An,p admits a representation p D �0 CPk

jD1 gj�j with �0; �j 2P

A2, �0 2 Al.n/ andgj�j 2 Al.n/ for j D 1; : : : ; k.

Indeed, suppose that Q is stable. Applying Definition 13.41 to V D An, then WV

is contained in Al.n/ for some l.n/ 2 N. Then, by (13.23), (�) holds. Conversely,assume that (�) is satisfied. Then, given V we choose n such that V An. Then (�)implies that V \Q P .Al.n/I g1; : : : ; gk/, so Q is stable.

Example 13.44 The simplest example of a stable quadratic module is the preorder-ing

PRdŒ x �2 of the polynomial algebra A D RdŒ x �.

In this case, g1 D 1; k D 1 and RdŒ x �2n \PRdŒ x �2 D PRdŒ x �2n by (13.1).

Thus condition (�) is satisfied for An D RdŒ x �n and l.n/ D n, soP

RdŒ x �2 isstable. ıExample 13.45 Let k 2 f1; : : : ; dg and m1; : : : ;mk 2 N. The quadratic module

Q DX

RdŒ x �2 C xm11

XRdŒ x �

2 C � � � C xmkk

XRdŒ x �

2

of A D RdŒ x � is stable. Indeed, setting An D RdŒ x �n one easily checks thatcondition (�) is fulfilled with l.n/ D nCmaxj mj. ı

The following proposition contains the crucial technical part of the proof ofTheorem 13.47 below, but it is also of interest in itself.


Proposition 13.46 Let Q be the quadratic module of A generated by g1; : : : ; gk.Suppose that Q\ .�Q/ D f0g and (13.25) holds. Let n D .n0; : : : nk/ 2 NkC1

0 . Then

Qn WDX

.An0 /2 C g1

X.An1 /

2 C � � � C gkX

.Ank/2 (13.26)

is a closed subset of Am for some m 2 N:

Proof Put g0 D 1. Since Al is finite-dimensional, each giP.Ai/

2 is contained insome finite-dimensional subspace of A, say of dimension r.i/. HenceQn is containedin a finite-dimensional subspace of A, so by (13.25) there is an m 2 N such thatQn Am. By Carathéodory’s theorem A.35, each element of the cone gi

P.Ai/

2 isa sum of r.i/ summands gjb2, where b 2 Ai. Thus Qn is the image of the map

˚ W Rn WD .An0 /r.n0/ � � � � � .Ank/

r.nk/ ! Am;

˚.. f0j/r.n0/jD1 � � � � � . fkj/r.nk/jD1 / WD

kXlD0

r.nk/XjD1

glf2lj :

We equip the finite-dimensional real vector spaces Rn and Am with some norm.Let S be the unit sphere of Rn. Since all linear and all bilinear mappings of finite-dimensional normed spaces are continuous,˚ W Rn ! Am is continuous. Hence theimage U WD ˚.S/ of the compact subset S of Rn is compact in Am.

Next we show that ˚ is injective. Assume that ˚.v/ D 0 for v 2 Rn. We write

˚.v/ DkX

lD0

r.nk/XjD1

gl f2lj

with f 2lj 2 Anj . Since ˚.v/ D 0, we have

g0f201 D �

r.n0/XjD2

g0f20j �

kXlD1

r.nk/XjD1

gl f2lj 2 Q\ .�Q/:

Since Q \ .�Q/ D f0g by assumption, g0f 201 D 0. Continuing this procedure byinduction it follows that glf 2lj D 0 for all l; j. Thus v D 0 and ˚ is injective.

Now we prove that Qn is closed in Am. Let . fn/n2N be a sequence of Qn whichconverges to some element f 2 Am. We will show that f 2 Qn: By definition wehave fn D ˚.vn/ for some vn 2 Rn. Writing vn D �nwn with wn 2 S and �n � 0,we get fn D �nun with un 2 ˚.S/ D U. Since U is compact in Am, the sequence.un/n2N has a subsequence which converges to some element u 2 U. For notationalsimplicity we assume that the sequence .un/ itself converges to u. Because ˚ is


injective, 0 … U and therefore u ¤ 0. Then limn �n D limnk fnkkunk D

k fkkuk and hence

f D limn

fn D limn�nun D .lim

n�n/.lim

nun/ D k fkkuk u 2

k fkkuk � U Qn: ut

The next theorem is our main result about the closedness of quadratic modules.

Theorem 13.47 Let A be a finitely generated commutative real unital algebra andQ a finitely generated quadratic module of A. Suppose that Q\ .�Q/ D f0g and Qis stable. Then Q is closed in the finest locally convex topology of A.

Proof Since A is finitely generated, we can write A in the form (13.25). In the casen WD n0 D � � � D nk the set Qn in (13.26) is just the cone

P.AnI g1; : : : ; gk/

for W D An defined by (13.23). Since Qn D P.AnI g1; : : : ; gk/ is closed by

Proposition 13.46, the assertion follows from Proposition 13.43. utThe assumption Q \ .�Q/ D f0g is fulfilled in the following important case.

Lemma 13.48 If Q is a quadratic module of RdŒ x � such that K.Q/ contains aninterior point, then Q \ .�Q/ D f0g.Proof Let p 2 Q \ .�Q/. Then p � 0 and �p � 0 on K.Q/. Therefore, thepolynomial p vanishes on K.Q/ and so on an open set. Hence p D 0. utCorollary 13.49 Let k 2 f1; : : : ; dg and m1; : : : ;mk 2 N. The quadratic modules

XRdŒ x �

2 C xm11X

RdŒ x �2 C � � � C xmk

k

XRdŒ x �

2

andP

RdŒ x �2 are stable and closed in RdŒ x �.

Proof Let Q be one of these quadratic modules. By Examples 13.44 and 13.45, Qis stable. Obviously, the positive d-octant .RC/d is a subset of K.Q/, so K.Q/ hasan interior point. Therefore Q is closed by Theorem 13.47 and Lemma 13.48. ut

The following Propositions 13.50 and 13.51 and Example 13.52 contain resultsabout the failure of (MP) for certain quadratic modules.

Proposition 13.50 Let Q be a finitely generated quadratic module of RdŒ x � suchthat K.Q/ contains an interior point. Suppose that d � 2 and Q is stable. Then Qdoes not have (MP).

Proof Let g1; : : : ; gk be generators of Q. We can assume that none of the gj is thezero polynomial. Set g WD g1 � � �gk. Since K.Q/ has a non-empty interior U, gcannot vanish on the whole set U, so there is a point x0 2 U K.Q/ such thatg.x0/ ¤ 0: Upon translation we can assume that x0 D 0: Since x0 D 0 2 K.Q/ andgj.0/ ¤ 0, it follows that gj.0/ > 0 for j D 1; : : : ; k: Further, we set g0 D 1 andchoose a polynomial p 2 Pos.Rd/ such that p … PRdŒ x �2; for instance, we maytake the Motzkin polynomial given by (13.6).

For n 2 N define pn.x/ WD p.2nx/ . We prove that not all polynomials pn are inQ. Assume to the contrary that pn 2 Q for all n 2 N. Since deg. pn/ D deg. p/ and


Q is stable, there exist m 2 N and representations

pn.x/ D p.2nx/ DkX

jD0gj.x/�nj.x/;

where �nj 2PRdŒ x �2m. Rescaling the preceding we get

p.x/ DkX

jD0gj.2

�nx/�nj.2�nx/; n 2 N: (13.27)

Since gj.0/ > 0, there are " > 0 and a ball B K.Q/ around 0 such that gj.x/ � ",hence gj.2�nx/ � ", on B for all j and n. Then, since all polynomials in (13.27)are nonnegative on B and the left-hand side does not depend on n, the values of�nj.2

�nx/ are uniformly bounded on B, that is, supx2B supn;j �nj.2�nx/ <1. Clearly,

�nj.x/ WD �nj.2�nx/ 2PRdŒ x �2m RdŒ x �2m. The supremum over B defines a normon the finite-dimensional space RdŒ x �2m and the sequences .�nj/n2N of polynomials�nj 2 RdŒ x �2m are bounded with respect to this norm. Hence they have convergingsubsequences .�nlj/l2N for j D 0; : : : ; k. Their limits �j are also in

PRdŒ x �2m by

Proposition 13.46 (or Corollary 13.49). Passing to the limit nl ! 1 in (13.27)yields

p.x/ DkX

jD0gj.0/�j.x/:

Since �j 2 PRdŒ x �2 and gj.0/ > 0, p 2 PRdŒ x �2, which contradicts the choiceof p.

By the preceding we proved that pn … Q for some n. Since Q is stable, Q is closedby Theorem 13.47 and Lemma 13.48. Therefore, pn … Q. But pn 2 Pos.Rd/. ThusQ ¤ Pos.Rd/. Hence Q does not have (MP) by Proposition 13.9(ii). ut

Let us mention a consequence of the preceding result. Let f be a finite subset ofRdŒ x � such that the semi-algebraic set K.f/ is compact. Then, by Theorem 12.25,T.f/ has (SMP) and hence (MP). Therefore, if d � 2 and K.f/ contains an interiorpoint, Proposition 13.50 implies that the preordering T.f/ is not stable!

There is an even stronger result than Proposition 13.50, proved by C. Scheiderer[Sr1, Theorem 5.4]. We state it here without proof (for the dimension of a semi-algebraic set we refer to [BCRo, Corollary 2.8.9]):

Let Q be a finitely generated quadratic module of RdŒ x �. If the semi-algebraicset K.Q/ has dimension at least 2 and Q is stable, then (MP) fails.

That is, (MP) and stability exclude each other if the dimension is at least 2.

Proposition 13.51 Let Q be a finitely generated quadratic module of RdŒ x � suchthat the semi-algebraic set K.Q/ contains an open cone C. Then Q is stable andclosed. If d � 2, then Q does not obey (MP) in RdŒ x �.


Proof Upon translation we can assume without loss of generality that the vertex ofC is the origin. Let p; q 2 Pos.C/. We show that

max.deg. p/; deg.q// D deg. pC q/: (13.28)

This is obvious if deg. p/ ¤ deg.q/ and if p D 0 or q D 0. Hence we can assumethat n WD deg. p/ D deg.q/ and pq ¤ 0. Since C has nonempty interior and pq ¤ 0,there exists an x0 2 C such that pq.x0/ ¤ 0. Hence p.x0/ > 0 and q.x0/ > 0. Forall � > 0 we have �x0 2 C and hence p.�x0/ � 0 and q.�x0/ � 0. For large �this implies that pn.x0/ > 0 and qn.x0/ > 0, where pn and qn are the correspondinghomogeneous terms of degee n. Thus . pn C qn/.x0/ > 0 and hence pn C qn ¤ 0.That is, deg. pC q/ D n, which proves (13.28).

Let g1; : : : ; gk be a set of generators of Q. Suppose that f D Pj gj�j 2 Q with

�j 2 PRdŒ x �2. Setting p D gj�j and q D Pi¤j gi�i, then p and q are nonnegative

on C and hence deg.gj�j/ D deg. p/ � deg. pCq/ D deg. f / by (13.28). Therefore,condition (�) is satisfied with An D RdŒ x �n and l.n/ D n, so that Q is stable.

Since C is contained in the interior of K.Q/, Q is closed by Theorem 13.47 andLemma 13.48. If d � 2, then Q does not obey (MP) by Proposition 13.50. ut

We illustrate the preceding with a simple, but typical example.

Example 13.52 Suppose that d � 2. The quadratic modules Q0 WDPRdŒ x �2 and

Qk DX

RdŒ x �2 C x1

XRdŒ x �

2 C � � � C xkX

RdŒ x �2; k D 1; : : : ; d;

do not satisfy (MP). Indeed, these quadratic modules are stable and closed byCorollary 13.49. Further, since K.Qk/ contains an open cone, Qk does not obey(MP) by Proposition 13.51 (or likewise by Proposition 13.50).

Because Qk is closed in RdŒ x �, by the separation theorem for convex sets, eachpolynomial p0 2 Pos.Rd/nQk gives rise to a Qk-positive linear functional L onRdŒ x � such that L. p0/ < 0. Each such functional L is not a moment functional.Recall that the Motzkin polynomial pc; 0 < c � 3, defined by (13.6) is inPos.R2/nQ0: ı

13.9 The Moment Problem on Some Cubics

In this section, we illustrate (SMP) and (MP) for some cubics. We use onlyelementary computations and do not require results from the theory of algebraiccurves.

Throughout this section, we suppose that f 2 RŒx1; x2� is a polynomial of degree3 and f3 is its homogeneous part of degree 3. We denote by

Cf � Z. f / D f.x1; x2/ 2 R2 W f .x1; x2/ D 0g

13.9 The Moment Problem on Some Cubics 345

the plane real curve associated with f , by I the ideal generated by f and by OI theideal of all p 2 RŒx1; x2� which vanish on Cf . Then RŒCf � D RŒx1; x2�= OI is thealgebra of regular functions on the curve Cf , see Eq. (12.7). Set

RŒCf �C WD fp 2 RŒCf � W p.x/ � 0 for x 2 Cf g:

Proposition 13.53 If f3 has a nonreal zero, thenP

RŒCf �2 obeys (SMP).

Proof The assumption implies that f3 has a real zero z0 and two nonreal complexconjugate zeros. Upon translation we can asume that z0 D .0; 1/, so f3 has the form

f3.x1; x2/ D x1.ax21 C bx1x2 C cx22/; where a; b; c 2 R; 4ac > b2:

In particular, we have a > 0. Hence f has degree at most 2 in x2, so we can writef DP2

jD0 pj.x1/xj2, where pj 2 RŒx1� and deg. pj/ � 3 � j.

Let x1 2 R and assume that there exists an x2 2 R satisfying .x1; x2/ 2 Cf . Thismeans that the quadratic equation f .x1; x2/ D 0 in x2 has a real solution, so that

g.x1/ WD p1.x1/2 � 4p0.x1/p2.x1/ � 0:

The polynomial g.x1/ has degree at most 4 and its coefficient of x41 is b2� 4ac < 0:Hence there exists a � > 0 such that g.x1/ < 0 for jx1j > �. Therefore, if jx1j > �,there is no x2 2 R such that .x1; x2/ 2 Cf . That is, the curve Cf is contained in thestrip Œ��; �� R of R2.

Let f D f f ;�f g: Then Cf D Z. f / is the semi-algebraic set K.f/ and thepreorderingT.f/ is

PRŒx1; x2�2CI, see e.g. Example 12.4. Since Cf Œ��; ��R,

the preordering T.f/ has (MP) by Proposition 13.25. By I OI, each positivelinear functional L on RŒCf � D RŒx1; x2�= OI lifts to a positive linear functional QLon RŒx1; x2� that vanishes on I. Since T.f/ D P

RŒx1; x2�2 C I, QL is T.f/-positive.Because T.f/ has (MP), the functional QL, hence L, is given by a Radon measure �.Since QL vanishes on I, � is supported on Cf by Proposition 12.19. That is,

PRŒCf �

2

has (SMP). utExample 13.54 Suppose that f3 D x31 C x32. Then f3 has a nonreal zero, henceP

RŒCf �2 satisfies (SMP) by Proposition 13.53. Examples of this kind are the

Fermat curve f D x31 C x32 � 1 and the folium of Descartes f D x31 C x32 � 3x1x2(Fig. 13.1). ı

Proposition 13.55 Suppose that the polynomial f is of the form

f D x31 C a20x21 C 2a11x1x2 C a02x

22 C a10x1 C a01x2 C a00; where a02 ¤ 0:

ThenP

RŒCf �2 is closed in the finest locally convex topology of the vector space

RŒCf � and does not obey (MP).


Fig. 13.1 Left: Fermat curve, middle: folium of Descartes. Both have (SMP), see Example 13.54.Right: Neil’s parabola. It does not have (MP), see Example 13.56

Proof Replacing x1 by x1� 13a20, the coefficient of x21 becomes zero. Since a02 ¤ 0,

the coefficient of x2 vanishes after some translation. Then f has the form

f .x1; x2/ D �bx22 C x31 C ax1x2 C cx1 C d; a; b; c; d 2 R; b ¤ 0: (13.29)

From (13.29) it follows that each element of the quotient algebra RŒCf � is of theform

g D p.x1/C x2q.x1/; where p; q 2 RŒx1�: (13.30)

(It suffices to note that x22 can be eliminated by using (13.29), since b ¤ 0:)First we prove that the quadratic module

PRŒCf �

2 is stable. For n 2 12N, let Vn

denote the vector space of elements g 2 RŒCf �, where deg. p/ � n and deg.q/ �n � 1 in (13.30). Let gi D pi.x1/ C x2qi.x1/, i 2 J, be finitely many elements ofRŒCf �. In the algebra RŒCf � we compute

Xig2i D

Xi. p2i C 2x2piqi C x22q

2i /

DX

i

�p2i C 2x2piqi C b�1.x31 C ax1x2 C cx1 C d/q2i

�

DX

i

�p2i C b�1.x31 C cx1 C d/q2i /C x2

Xi

�2piqi C ab�1x1q2i

�

DW p.x1/C x2q.x1/: (13.31)

Suppose thatP

g2i 2 Vn, that is, deg. p/ � n and deg.q/ � n � 1. We shall showthat gi 2 Vn=2 for all i 2 J. Set j WD maxi deg. pi/ and k WD maxi deg.qi/.

Case 1: j > kC 1.Then 2j > 2k C 3 and hence 2j D deg. p/ � n by (13.31). Thus, deg. pi/ � j �

n=2 and deg.qi/ � k < j� 1 � n=2� 1, so that gi 2 Vn=2 for i 2 J:Case 2: j � kC 1.

13.9 The Moment Problem on Some Cubics 347

Then 2j < 2k C 3 and hence 2k C 3 D deg. p/ � n by (13.31). Therefore, wehave deg.qi/ � k < n=2� 1 and deg. pi/ � j � kC 1 < n=2; so that gi 2 Vn=2 fori 2 J.

The preceding proves that Q WD PRŒCf �

2 is stable. Further, Q \ .�Q/ D f0gby (12.8). It follows therefore from Theorem 13.47 that Q DPRŒCf �

2 is closed inRŒCf � with respect to the finest locally convex topology.

Put � WD sign b. We show that there is a real number ˛ such that

�x1 � ˛ 2 RŒCf �C and �x1 � ˛ …X

RŒCf �2: (13.32)

Let .x1; x2/ 2 Cf . Then �bx22 C ax1x2 C x31 C cx1 C d D 0. This is a quadraticequation in x2 which has a solution. Therefore,

a2x21 C 4b.x31 C cx1 C d/ � 0:

Suppose first that b > 0. Then limx1!�1 Œa2x21 C 4b.x31 C cx1 C d/� D �1, sothere is a real number ˛ such that a2x21 C 4b.x31 C cx1 C d/ < 0 for x1 < ˛. Hencex1 � ˛ for all .x1; x2/ 2 Cf , that is, �x1 � ˛ 2 RŒCf �C for b > 0. In the case b < 0

we replace x1 by �x1 and use the same reasoning.To prove that �x1�˛ …PRŒCf �

2 we assume to the contrary that �x1�˛ DP g2i ;where gi 2 RŒCf �

2. Since �x1 � ˛ 2 V1, it was shown above that all gi are in V1=2,so they are constants. From (13.29) it is obvious that x1 is not constant on Cf , so wehave a contradiction. This completes the proof of (13.32).

By (13.32) we have RŒCf �C ¤PRŒCf �

2. SinceP

RŒCf �2 is closed, as shown

above, it follows from Proposition 13.9(ii) (or directly from the separation theoremfor convex sets) that Q DPRŒCf �

2 does not obey (MP). utExample 13.56 Neil’s parabola (Fig. 13.1) f .x1; x2/ D x31 � x22.

Then x1 2 RŒCf �C and x1 … PRŒCf �2. Indeed, it is obvious that x1 � 0 on

Z. f /, that is, x1 2 RŒCf �C. Arguing as in the preceding proof of Proposition 13.55it follows that x1 …PRŒCf �

2:

By Proposition 13.55,P

RŒCf �2 does not satisfy (MP). In fact, any positive linear

functional L on RŒCf � such that L.x1g/ < 0 for some g 2PRŒCf �2 is not a moment

functional. Likewise a linear functional QL on RŒx1; x2� which vanishes on I andsatisfies QL.x1g/ < 0 for some g 2 PRŒx1; x2�2 cannot be a moment functional. InExercise 13.3 we use this observation to “construct” a positive linear functional onRŒx1; x2� which is not a moment functional. ı

Let us return to a general cubic. Since f3 is a homogeneous polynomial in twovariables of degree 3, it is a product of 3 linear factors. If one of the zeros of f3 isnonreal, or equivalently, f has a nonreal zero at infinity, then RŒCf � has (SMP) byProposition 13.53. If all zeros of f3 are real, it can happen that RŒCf � obeys (SMP)(see Example 13.57 with p.x1/ D x31) and that RŒCf � does not have (MP) as shownby Proposition 13.55.

The aim of this section was to give a small glimpse into the moment problem oncurves; this problem is extensively studied in [PoS, Sr2], and [Pl].


We close this section with a very simple example of curves which have (SMP).

Example 13.57 Let p 2 RŒx1; x2� and set g.x1; x2/ D x2 � p.x1/: We consider theplane curve Cg D Z.g/ and show that the preordering

PRŒCg�

2 has (SMP).Let q 2 RŒCg�C. Then q.x1; p.x1// 2 RŒx1� is nonnegative on R. Therefore,

by Proposition 3.1(i), q.x1; p.x1// D q1.x1/2 C q2.x1/2 for some q1; q2 2 RŒx1�, sothat q 2PRŒCg�

2. By Haviland’s theorem 1.14, applied to the algebra A D RŒCg�,each positive linear functional on RŒCg� is a moment functional with representingmeasure on OA D Cg. This means that

PRŒCg�

2 has (SMP). ı

13.10 Proofs of the Main Implications of Theorems 13.10and 13.12

In this section, we also use the polynomial algebra RmŒy� WD RŒy1; : : : ; ym�.Let us begin with two technical lemmas.

Lemma 13.58 Let �1 and �2 be Radon measures on a compact subset K of Rm

and let A be a Borel subset of K. Then there exists a subset N of K such that�1.N/ D �2.N/ D 0 and a sequence .qn/n2N of polynomials qn 2 RmŒy� suchthat lim

n!1 qn.�/ D �A.�/ for � 2 KnN and sup�2Kjqn.�/j � 2 for n 2 N.

Proof Clearly, it suffices to prove the assertion for the case when K is an m-dimensional compact interval. Since the measures are regular, there exist compactsubsets C0

n;C00n and open subsets U0

n;U00n of K such that

C0n A U0

n; C00n A U00

n ; �1.U0nnC0

n/ � 2�n; �2.U00n nC00

n / � 2�n for n 2 N:

Then Cn A Un, �1.UnnCn/ � 2�n and �2.UnnCn/ � 2�n for the compact setCn D C0

n [ C00n and the open set Un D U0

n \ U00n . Further, without loss of generality

we can assume Cn CnC1 and UnC1 Un, so that UnC1nCnC1 UnnCn. ApplyingUrysohn’s lemma to the compact set Cn and the closed set RmnUn of Rm there areexists a continuous function 'n on K with values in Œ0; 1� such that 'n D 1 onCn and 'n D 0 on RmnUn. From Weierstrass’ theorem, applied to the compact m-dimensional interval K, it follows that there exists a polynomial qn 2 RmŒy� suchthat jqn.�/ � 'n.�/j � 2�n for � 2 K. Clearly, we have sup�2K jqn.�/j � 2: SetN D \nUnnCn. Then �j.N/ D 0 for j D 1; 2, since �j.\nUnnCn/ � �j.UknCk/ �2�k for all k 2 N.

Let � 2 KnN. Then � … UknCk for some k. For n � k we have UnnCn UknCk,so that � … UnnCn and hence 'n.�/ D �A.�/. Thus limn 'n.�/ D �A.�/. Sincelimn.qn.�/ � 'n.�// D 0, we conclude that limn qn.�/ D �A.�/. utLemma 13.59 Suppose that � is a Radon measure on a compact subset K ofRm and ' is a real-valued function of L1.KI�/. If RK q.�/'.�/d�.�/ � 0 (resp.RK q.�/'.�/d�.�/ D 0) for all q 2 RmŒy�, then ' � 0 (resp. ' D 0) �-a.e. on K.

13.10 Proofs of the Main Implications of Theorems 13.10 and 13.12 349

Proof Given " > 0, we put A" WD f� 2 K W '.�/ � �"g. For the Borel set A" wechoose a sequence .qn/n2N of polynomials as in Lemma 13.58. Since jq2n'j � 4j'jon K and ' 2 L1.KI�/, Lebesgue’s dominated convergence theorem A.2 appliesand yields limn

Rq2n'd� D

R�A"'d�. Hence

R�A"'d� � 0, since

Rq2n'd� � 0

by assumption. On the other hand,R�A"'d� � �"�.A"/ by the definition of A".

Therefore, �.A"/ D 0. Since " > 0 was arbitrary, '.�/ � 0 �-a.e. on K.The second assertion follows by applying the preceding to ' and �'. utAs noted in Sect. 13.3 the proofs of Theorem 13.10 and 13.12 are complete once

the main implications (ii)!(i) are proved.

Proof of the Implication (ii)!(i) of Theorem 13.12 Let ˛j and ˇj be as inTheorem 13.12. Throughout this proof we abbreviate

K DmYjD1Œ˛j; ˇj�:

Let us fix a Q-positive linear functional L on A. For any p 2 A we define a linearfunctional Lp on the polynomial algebra RmŒy� by

Lp.q/ D L.q.h/p2/; q 2 RmŒy�: (13.33)

Note that q.h/ WD q.h1; : : : ; hm/ is a well-defined element of A, since h1; : : : ; hm 2 Aby assumption.

Lemma 13.60 For p 2 A there exists a Radon measure � on K such that

Lp.q/ D L.q.h/p2/ DZKq d�p; q 2 RmŒy�: (13.34)

Proof The moment problem for the m-dimensional interval K was treated inExample 12.47. As shown therein (see (12.38)), for the existence of a measure �psuch that (13.34) holds it suffices to verify that for all q 2 RmŒy� and j D 1; : : : ;m,

Lp..ˇj � yj/q2/ � 0 and Lp..yj � ˛j/q2/ � 0: (13.35)

By assumption, ˇj � hj belongs to Q. Because Q is also a quadratic module (as Qis), .ˇj � hj/q.h/2p2 2 Q. Therefore, since L is Q-positive by assumption and henceQ-positive, using (13.33) we obtain

Lp..ˇj � �j/q2/ D L..ˇj � hj/q.h/2p2/ � 0

which implies the first equality of (13.35). The second one is derived in a similarmanner from the assumption .hj � ˛j/ 2 Q. ut


Since K is compact, the measure �p is uniquely determined by p, but we willnot need this here. Let � denote the measure �p obtained for the element p D 1. Acrucial step of this proof is contained in the following

Lemma 13.61 For all p 2 A the measure �p is absolutely continuous with respectto �, that is, each �-null set is also a �p-null set.

Proof Asssume that A is a �-null set. For the set A, the compact set K and themeasures �1 D �p, �2 D � we choose a sequence .qn/n2N of polynomials qn 2RmŒy� as in Lemma 13.58. Then we have limn qn.�/ D �A.�/ �p-a.e. on K andlimn qn.�/2 D �2A.�/ D �A.�/ �-a.e. on K. Therefore, by Lebesgue’s dominatedconvergence theorem A.2, we have

limn

ZKqnd�p D

ZK�Ad�p D �p.A/ and lim

n

ZKq2nd� D

ZK�Ad� D �.A/ D 0:

(13.36)

Further, sinceP

A2 Q and L.Q/ � 0, L is a positive functional on A, so theCauchy–Schwarz inequality (2.4) holds. From this inequality and (13.34) we obtain

�ZKqn d�p

�2D L

�qn.h/p

2�2 � L

�qn.h/

2�L�p4� D L. p4/

ZKq2n d�: (13.37)

Combining (13.36) and (13.37) we conclude that �p.A/ D 0. utBy Lemma 13.61, the assumptions of the Radon–Nikodym theorem A.3 are

fulfilled, so there exists a nonnegative function �p 2 L1.K; �/ such that d�p D �pd�.Then, by (13.34) we have

Lp.q/ D L.q.h/p2/ DZKq�p d�; q 2 RmŒy�: (13.38)

Now we take a basis of the form fp2j gj2N, where N N; of the real vector spaceA. (Such a basis exists: Since the algebra A is finitely generated, its vector spacedimension is at most countable. If f flg is a vector space basis of A, we may takeelements fl ˙ 1 as pj. These elements span A, since . fl C 1/2 � . fl � 1/2 D 4fl.)

For � 2 K we define a linear functional L� on A by

L�

�Xjcjp

2j

�DX

jcj�pj.�/: (13.39)

Then we have the following integral decomposition of the functional L:

L.q.h/p/ DZKq.�/L�. p/ d�.�/ for p 2 A; q 2 RmŒy�: (13.40)

13.10 Proofs of the Main Implications of Theorems 13.10 and 13.12 351

Indeed, using (13.38) and (13.39), for p DPj cjp2j we prove (13.40) by computing

L.q.h/p/ DXj

cjL.q.h/p2j / D

Xj

cj

ZKq.�/�pj.�/ d�.�/

DZKq.�/

�Xj

cj�pj.�/

�d�.�/ D

ZKq.�/L�. p/ d�.�/:

The next technical steps are collected in the following lemma.

Lemma 13.62

(i) L�. p/ � 0 �-a.e. on K for any p 2 Q.(ii) L�..hj � �j/p/ D 0 �-a.e. on K for p 2 A and j D 1; : : : ;m.

(iii) There exists a �-null set N such that L� � 0 on Qg.�/ for � 2 KnN.Proof

(i) Let p 2 Q. Then q.h/2p belongs to the quadratic module Q for any q 2 RmŒy�and hence

L.q.h/2p/ DZKq.�/2L�. p/ d�.�/ � 0

by (13.40). Therefore, since the function � 7! L�. p/ is L1.K; �/ (because thefunctions �pj are), Lemma 13.59 applies and yields L�. p/ � 0 �-a.e. on K.

(ii) Let q 2 RmŒy�. Applying formula (13.40) twice we derive

ZKq.�/L�..hj��j/p/ d�.�/ D L.q.h/.hj��j/p/ D

ZKq.�/.�j��j/L�. p d�.�/ D 0:

Using again Lemma 13.59 it follows that L�..hj��j/p/ D 0 �-a.e. on K.(iii) Let g1; : : : ; gk be generators of the finitely generated quadratic moduleQ. From

(i) and (ii) it follows that for the countable (!) subsets

Q0 DX

QŒ x �2 C g1X

QŒ x �2 C � � � C glX

QŒ x �2;

I 0� D .h1 � �1/QŒ x �C � � � C .hm � �m/QŒ x �; � 2 K;

of Q and I�, respectively, there exists a common �-null set N such that forall � 2 KnN we have L�. p/ � 0 for p 2 Q0 and L�.r/ D 0 for r 2 I 0

�.Approximating real coefficients by rationals it follows that this holds for allelements p 2 Q and r 2 I� as well. Therefore, for � 2 KnN, we have L� � 0on Q and L� D 0 on I�. Thus L� � 0 on Q� D Q C I� and hence on theclosure Q�. ut

Now we can complete the proof of the implication (ii)!(i) of Theorem 13.12.Suppose that Q� has (SMP) for all � 2 K. Recall that L is an arbitrary Q-positive


linear functional on A. Let p 2 Pos.K.Q//: Then, obviously, p 2 Pos.K.Q�// for� 2 K. Since Q� has (SMP), Q� D Pos.K.Q�// by Proposition 13.9(i). Therefore,p 2 Q�. Therefore L�. p/ � 0 �-a.e. on K by Lemma 13.62(iii) and hence L. p/ � 0by (13.40). That is, L � 0 on Pos.K.Q//: Hence Haviland’s Theorem 1.14 (i)!(iv),applies and shows that L comes from a measure � 2 MC. OA/ supported on K.Q/.This proves that Q has (SMP).

The arguments in the case of (MP) are almost the same. By Proposition 13.9(iii),it suffices to replace in the preceding paragraph Pos.K.Q// and Pos.K.Q�// byPos. OA/. Now the proof of Theorem 13.12 is complete. �

Remark 13.63 In fact, we have shown under the assumptions of Theorem 13.12 that

Q D \�2QmjD1 Œ˛j;ˇj�

Q�:

Indeed, since Q Q� by definition, the inclusion Q \�Q� is obvious.Conversely, let p 2 \�Q�. Then, as shown at the end of the preceding proof,L. p/ � 0 for each Q-positive linear functional L. Hence p 2 Q by Lemma 13.8. ıProof of the Implication (ii)!(i) of Theorem 13.10 In order to apply Proposi-tion 12.23(ii) we first note that we can assume without loss of generality thatA D RdŒ x �. Recall that A is (isomorphic to) the quotient algebra RdŒ x �=J forsome ideal J of RdŒ x � and � W RdŒ x � ! A denotes the canonical map. Wechoose polynomials Qhj 2 RdŒ x � such that �.Qhj/ D hj. Clearly, QT WD ��1.T/ is apreordering of RdŒ x �. Since J QT , each QT-positive character of RdŒ x � annihilatesJ , so it belongs to OA D Z.J /. This implies that K. QT/ D K.T/ and Qhj.x/ D hj.x/for x 2 K.T/: Thus Qh.K. QT// D h.K.T//: Since hj is bounded on K.T/ by (ii),so is Qhj on K. QT/. For � 2 h.K.T//; QI� WD ��1.I�/ is an ideal of RdŒ x �. It iseasily verified that . QT/� D QT C QI�: Since QT D QT C J and . QT C J /=J Š T, byProposition 13.14(iii), QT has (SMP) resp. (MP) in RdŒ x � if and only if T has in A.The same holds for T�. This shows that we can assume that A D RdŒ x �.

By the assumption of Theorem 13.10, each element hj, j D 1; : : : ;m, of RdŒ x � isbounded on K.T/, that is, there are numbers ˛j; ˇj 2 R such that ˛j � hj.x/ � ˇjon K.T/. Since ˇj�hj and hj�˛j are bounded and nonnegative on K.T/, we haveL.ˇj� hj/ � 0 and L.hj�˛j/ � 0 for each T-positive linear functional L on RdŒ x �by Proposition 12.23(ii). Therefore, by Lemma 13.8, ˇj�hj and hj�˛j belong to theclosure T of the preordering T. Thus the preordering T satisfies the assumptions ofTheorem 13.12. Hence, by Theorem 13.12, T has (SMP) resp. (MP) provided thatT� has (SMP) resp. (MP) for all � 2 K D Qm

jD1 Œ˛j; ˇj�: To complete the proof ofTheorem 13.10 it remains to show that it suffices to assume this for � in the smallerset h.K.T//:

Fix a T-positive linear functional L on RdŒ x �. We show that the measure � D �1for the polynomial p D 1 in Lemma 13.60 is supported on the set H WD h.K.T// .Suppose that q 2 RmŒy� is nonnegative on H. Then q.h/ 2 RdŒ x � is nonnegative onK.T/. Further, q.h/ is bounded on K.T/, because the polynomials hj are. Therefore,since L is T-positive, we have L1.q/ D L.q.h// � 0 again by Proposition 12.23(ii).

13.11 Exercises 353

This shows that the assumptions of Haviland’s Theorem 1.12 for the closed set Hare satisfied. Hence, by this theorem, supp � H. Consequently, the integration informula (13.40) is over H rather than K. Continuing along the lines of the aboveproof of Theorem 13.12 we see that it suffices to have (SMP) for � 2 H D h.K.T//:

Finally, let � 2 Hnh.K.T//. From its definition it follows that the fibre set K.T/�is empty, so that �1 2 T� by Theorem 12.3(iv) and hence RdŒ x � D T� D T� .Therefore, the requirement p 2 T� for p 2 Pos.K.T// (see the paragraph beforelast in the above proof of Theorem 13.12) is always fulfilled. That is, it suffices toassume (SMP) for � 2 h.K.T//.

The reasoning for (MP) is similar. The proof of Theorem 13.10 is complete. �

Remark 13.64 Under the assumption of Theorem 13.10 we proved that

T D \�2h.K.T// T�: ı

13.11 Exercises

1. Show that the Choi–Lam polynomial

p.x1; x2; x3/ WD x21x22 C x21x

23 C x22x

23 C 1 � 4x1x2x3 (13.41)

is nonnegative on R3, but it is not a sum of squares in RŒx1; x2; x3�.2. Use the Gram matrix representation (13.3) to prove that

PRdŒ x �2n is a closed

subset of RdŒ x �2n.3. (A positive functional on RŒx1; x2� that is not a moment functional [Sm7])

Let L be a T.x3/-positive linear functional on RŒx� that is not T.x/-positive. (Such a functional L was constructed in Example 13.7). Define linearfunctionals L1 on RŒx� and L2 on RŒx1; x2� by L1.x2n/ D L.xn/;L1.x2nC1/ D 0

for n 2 N0 and L2. p/ D L1. p.x2; x3// for p 2 RŒx1; x2�, that is, L2.xk1x2l2 / D

L.xkC3l/ and L2.xk1x2lC12 / D 0 for k; l 2 N0:

a. Show that L2 is a positive linear functional on RŒx1; x2�.Hint: Use the T.x3/-positivity of L.

b. Show that L2 is not a moment functional.Hint: Show that any representing measure would be supported on the curvex31 D x22. This is impossible, since L is not T.x/-positive. See Example 13.56.

4. Construct positive linear functionals on the preorderings of Examples 13.27and 13.28 which are not K.f/-moment functionals.

5. Modify the set of polynomials in Example 13.27 such that the set K.f/ remainsthe same, but the preordering T.f/ satisfies (SMP).

6. Let E 2 D.RŒx2�/ be finitely generated, p a nonconstant polynomial from RŒx1�and ˛ > 0. Suppose that D 2 D.RŒx1; x2�/ is generated by E and p2C˛. Showthat the preordering

P.AD/2 obeys (MP) in the algebra AD .


7. Decide whether or not the following preorderings satisfy (SMP).

a. f1.x/ D x1; f2.x/ D 1 � x1; f3.x/ D x1x2; f4.x/ D 1 � x1x2:b. f1.x/ D x1; f2.x/ D 1 � x1; f3.x/ D x2; f4.x/ D 1 � x1x2:c. f1.x/ D x1; f2.x/ D 1 � x1; f3.x/ D x32; f4.x/ D 1 � x1x2:

8. Set f D fx1; x2; 1 � x1x2; x1x2 � 1g and let I be the ideal generated by x1x2 inRŒx1; x2�.

a. Does the quadratic module I of RŒx1; x2� satisfy (SMP) or (MP)?b. Does the preordering T.f/ of RŒx1; x2� satisfy (SMP) or (MP)?

9. Show that the preordering T.x1; x2/ of RŒx1; x2� is closed and it does not have(MP).

10. Decide whether or not the preorderingP

RŒCf � of the algebra RŒCf � obeys(SMP) resp. (MP):

a. f .x1; x2/ D x31 C x21x2 C x1x22 C x32 C x1x2 C 1.b. f .x1; x2/ D x31.c. f .x1; x2/ D x31 � 2x22.d. f .x;x2/ D x31 C x1x22 � x22 (cissoid of Diocles).

11. (Moment problem on a one point set of R )Let A be the quotient algebra of RŒx� by the ideal generated by x2.

a. Show that OA D f�0g, where �0 is the point evaluation at zero.b. Show that

PA2 is not closed and that its closure is AC.

c. Show thatP

A2 has (SMP).d. What happens with assertions a.-c. if A is replaced by the quotient algebra

of RŒx� by the ideal generated by x?

12. ( Moment problem on two intersecting lines in R2)Consider two lines in R2 intersecting in one point and given by equations

a1x1 C b1x2 C c1 D 0 and a2x1 C b2x2 C c2 D 0: Let A be the quotient algebraof RŒx1; x2� by the ideal generated by .a1x1C b1x2C c1/.a2x1C b2x2C c2/:

a. Show that for each f 2 AC there exist g1; g2 2 A such that f D g21 C g22.b. Conclude that

PA2 has (SMP).

13. What happens if the two lines in Exercise 12 are parallel?14. Let p2 2 RŒx1�; p3 2 RŒx1; x2�; : : : ; pd 2 RŒx1; : : : ; xd�1�, where d � 2:

Consider the curve V in Rd given by the equations

xd D pd.x1; : : : ; xd�1/; : : : ; x3 D p3.x1; x2/; x2 D p2.x1/:

Show thatP

RŒV�2 has (SMP).15. a. Show that the convex hull of a compact subset of Rn is also compact.

b. Suppose that M is a convex compact subset of Rn such that 0 … M. Showthat the cone generated by M is closed in Rn.

13.12 Notes 355

c. Show by an example that the assertion of b. does not hold in general if theassumption 0 … M is omitted.

16. Let A be a finitely generated (real or complex) unital �-algebra and let q bea seminorm on A such that q.1/ D 1, q. fg/ � q. f /q.g/ and q. f �/ D q. f /for f ; g 2 A. Let L be a positive linear functional on A and let �L denote theGNS representation associated with L (see Definition 12.39). Suppose that thereexists a constant c > 0 such that jL. f /j � cq. f / for all f 2 A:

a. Use the Cauchy–Schwarz inequality to show that jL. f /j � L.1/q. f / forf 2 A.

b. Show that �L. f / is bounded and k�L. f /k � q. f / for f 2 A.c. Show that there exists a compact subset K of OA and a Radon measure � on

K such that L. f / D RK f .x/d�.x/ for f 2 A.

13.12 Notes

The existence of positive functionals on RdŒ x �, d � 2, which are not momentfunctionals was first shown rigorously in [Sm2] and [BCJ]. The explicit functionalin Proposition 13.5 is due to J. Friedrich [Fr1]. Hilbert’s method of constructingpositive polynomials that are not sos is elaborated in [Re4].

The fibre theorem was discovered by the author in [Sm7] for A D RdŒ x �; theproperties (SMP) and (MP) were also invented therein. The extended and moregeneral form stated as Theorem 13.10 was obtained only recently in [Sm11]. Theoriginal proof in [Sm7] was based on the decomposition theory of states [Sm4].An “elementary” proof was given by T. Netzer [Nt], see also [Ms1, Section 4.4].Our approach is based on arguments from [Nt] and [Ms1]. The weaker version forquadratic modules is taken from [Ms1]. Proposition 13.14 was noted in [Sr1].

The results of Sect. 13.4 are due to S. Kuhlmann and M. Marshall [KM], see also[KMS]. Proposition 13.25 was proved in [Mt, Theorem 1.3] and later independentlyin [KM]; the elegant approach given in the text is from [Sm7]. The applications ofthe fibre theorem to the rational moment problem in Sect. 13.6 and to the extensiontheorem in Sect. 13.7 are taken from [Sm11]. Further results on rational momentproblems can be found in [Ch, BGHN] (dimension one) and [PVs1, CMN, Sm11](multidimensional case).

That the coneP

RdŒ x �2 is closed in the finest locally convex topology was firstshown in [Sm1] and independently in [BCJ] ; a general result of this kind wasgiven in [Sm4, Theorem 11.6.3]. The stability concept of quadratic modules andits applications to the failure of (MP) in Sect. 13.8 are due to C. Scheiderer, see[PoS, Sr1]. The proof of Proposition 13.46 is taken from [Sw3]. Proposition 13.50appeared in [Sr1], while Proposition 13.51 is from [Ms1].

The moment problem on curves was first studied in [Mt] and [St1]. In Sect. 13.9we followed J. Matzke’s thesis [Mt], where the case of cubics was completelysettled. The moment problem on curves is now well-understood, see [Sr2, Pl].

Chapter 14The Multidimensional Moment Problem:Determinacy

In this chapter, we study the determinacy problem in the multivariate case. InSect. 14.1, we introduce several natural determinacy notions (strict determinacy,strong determinacy, ultradeterminacy) that are all equivalent to the “usual” deter-minacy in dimension one. In the remaining sections we develop various techniquesand methods to derive sufficient criteria for determinacy. In Sect. 14.2, polynomialapproximation is used to show that the determinacy of all marginal sequencesimplies the determinacy of a moment sequence (Theorem 14.6). Section 14.3 isbased on operator-theoretic methods in Hilbert space. The main results (Theo-rems 14.12 and 14.16) show that the determinacy of appropriate 1-subsequencesof a positive semidefinite d-sequence s implies that s is a (determinate) momentsequence. Section 14.4 is concerned with Carleman’s condition in the multivariatecase. Probably the most useful result in this chapter is Theorem 14.20; it saysthat if all marginal sequences of a positive semidefinite d-sequence s satisfyCarleman’s condition, then s is a determinate moment sequence. Section 14.6 usesthe disintegration of measures as a powerful method for the study of determinacy. Afibre theorem for determinacy (Theorem 14.30) states a measure is determinate if thebase measure is strictly determinate and almost all fibre measures are determinate.In Sect. 14.5, we calculate the moments of the surface measure on Sd�1 and of theGaussian measure on Rd.

14.1 Various Notions of Determinacy

First let us recall some results from the one-dimensional case. If � 2MC.R/ is ameasure with moment sequence s, we know from Theorem 6.10 and Corollary 6.11that the following statements are equivalent:

(i) The measure �, or equivalently, its moment sequence s, is determinate.(ii) The multiplication operator X by the variable x with domainCŒx� is essentially

self-adjoint on the Hilbert space L2.R; �/.


357

358 14 The Multidimensional Moment Problem: Determinacy

(iii) CŒx� is dense in L2.R; .1C x2/d�/.Further, if � is determinate, then the polynomials CŒx� are dense in L2.R; �/.

In dimensions d � 2 determinacy is much more subtle and the precedingconditions lead to different determinacy notions.

Recall that MC.Rd/ is the set of Radon measures on Rd for which all momentsare finite. For � 2MC.Rd/ we denote by M� the set of measures � 2MC.Rd/

which have the same moments as �, or equivalently, which satisfy

ZRd

p.x/ d�.x/ DZRd

p.x/ d�.x/ for all p 2 RdŒ x �:

We write � Š � if � 2M�. Obviously, “Š” is an equivalence relation in MC.Rd/.

Definition 14.1 For a measure � 2MC.Rd/ we shall say that

� � is determinate if M� is a singleton, that is, if � 2M� implies � D �,� � is strictly determinate if � is determinate and CdŒ x � is dense in L2.Rd; �/,� � is strongly determinate if CdŒ x � is dense in L2.Rd; .1Cx2j /d�/ for

jD1; : : : ; d,� � is ultradeterminate if CdŒ x � is dense in L2.Rd; .1C kxk2/d�/.In this section we adopt the following notational convention: If � is a measure

on Rd and no confusion can arise, we often write Lp.�/ instead of Lp.Rd; �/.Suppose that � 2 MC.Rd/: Let s be the moment sequence of � and L the

moment functional of �, that is, L. f / D Rf .x/d�; f 2 CdŒ x �. As in the one-

dimensional case the above notions will be used synonymously for �, s; and L.That is, we say that s, and likewise L, is determinate, strongly determinate, strictlydeterminate, ultradeterminate, if � has this property. Let us set Ms DML DM�.

We recall the Hilbert space approach and the GNS construction from Sect. 12.5.Let L be a positive linear functional on CdŒ x �. Then there is a scalar product h�; �iLon the vector space DL D CdŒ x �=NL such that h p C NL; q C NLiL D L. pq/,p; q 2 CdŒ x �; where NL D f f 2 CdŒ x � W L. f f / D 0g. Further, there are pairwisecommuting symmetric operators Xj, j D 1; : : : ; d, defined by Xj. pCNL/ D xjpCNL, p 2 CdŒ x �. The Hilbert space completion of the unitary space DL is denoted byHL. For notational simplicity we write p for pCNL. Then

h p; qiL D L. pq/ and Xjp D xjp; p; q 2 CdŒ x �;

that is, Xj is the multiplication operator by the variable xj: Recall that Xj denotes theclosure of operator Xj.

All four determinacy notions have been defined in terms of the measure �.By Theorem 14.2 below, � is strongly determinate if and only if the closures ofall symmetric operators X1; : : : ;Xd are self-adjoint. Thus, strong determinacy is anatural and fundamental concept from the operator-theoretic point of view.

14.1 Various Notions of Determinacy 359

Our main reason for introducing strict determinacy is that it enters into thefibre Theorem 14.30 below. Another reason stems from the theory of orthogonalpolynomials: Since the polynomials are dense in L2.Rd; �/, we have HL ŠL2.Rd; �/. Hence for a strictly determinate measure � each sequence of orthonor-mal polynomials in HL is an orthonormal basis of L2.Rd; �/.

By definition strict determinacy implies determinacy. Theorem 14.2 below showsthat if � is strongly determinate it is strictly determinate and hence determinate.Since the norm of L2.Rd; .1 C jjxjj2/d�/ is obviously stronger than that ofL2.Rd; .1C x2j /d�/, ultradeterminacy implies strong determinacy. Thus we have

ultradeterminate) strongly determinate) strictly determinate) determinate:

From Weierstrass’ approximation theorem it follows that each measure withcompact support is ultradeterminate and hence all four determinacy notions arefulfilled.

As noted at the beginning of this section, in the case d D 1 the fourconcepts (determinacy, strict determinacy, strong determinacy, ultradeterminacy)are equivalent. However, in dimension d � 2 all of them are different! An exampleof a strongly determinate measure that is not ultradeterminate is sketched in Exercise14.6. A strictly determinate measure which is not strongly determinate is developedin Example 14.4. In [BT] it is shown that there exist rotation invariant determinatemeasures � 2MC.Rd/, d � 2, such that CdŒ x � is not dense in L2.Rd; �/. Suchmeasures are not strictly determinate.

The following theorem gives an operator-theoretic characterization of strongdeterminacy.

Theorem 14.2 Let L be a moment functional on CdŒ x � and let � be a representingmeasure for L. Then � is strongly determinate if and only if all symmetric operatorsX1; : : : ;Xd are essentially self-adjoint on CdŒ x � in HL, that is, we have Xk D .Xk/

�for k D 1; : : : ; d. If this holds, then� is strictly determinate, that is,� is determinateand CdŒ x � is dense in L2.Rd; �/.

The following technical lemma is needed in the proof of Theorem 14.2.

Lemma 14.3 Suppose A is a closed symmetric operator on a Hilbert space K. LetH be a closed subspace of K and let P denote the orthogonal projection onto H.Suppose that D is a dense linear subspace of H such that D D.A/, AD 2 H andX WD AdD is an essentially self-adjoint operator on H. Then PA AP. Moreover,if Y denotes the restriction of A to .I � P/D.A/, then A D X˚ Y on K D H˚H?.

Proof Suppose that ' 2 D.A/. Let 2 D. Using the facts that and X are in Hand the relation X A it follows that

hX ;P'i D h pX ; 'i D hX ; 'i D hA ; 'i D h ;A'i D h p ;A'i D h ;PA'i:

Since 2 D was arbitrary, P' 2 D.X�/ and X�P' D AP'. Since X is essentiallyself-adjoint, X� D X and hence AP' D XP' D PA'. That is, PA AP.


If ' 2 D.A/, we have P' 2 D.X/ and .I � P/' 2 D.A/ as just shown. Hence' D P' C .I � P/' 2 D.X/˚D.Y/ and A' D X' C Y.I � P/' which proves thatA X ˚ Y. By definition the converse inclusion is clear. utProof of Theorem 14.2 The proof is based on the Hilbert space approach given byTheorem 12.40.

Suppose first that the symmetric operators X1; : : : ;Xd are essentially self-adjointon HL. Let .A1; : : : ;Ad/ be a d-tuple of strongly commuting self-adjoint operatorson a larger Hilbert space K such that Xj Aj, jD1; : : : ; d. By Lemma 14.3 thereis a decomposition Aj D Xj ˚ Yj on K D HL ˚ H?

L . Since Aj is self-adjoint,so is Yj and for the spectral measures we obtain EAj D EXj

˚ EYj : Hence the

spectral measures of Xk and Xl commute and (15.6) implies that E.X1;:::;Xd/

E.A1;:::;Ad/ for the corresponding joint spectral measures. Therefore, the measure�.�/ WD hE.X1;:::;Xd/

.�/1; 1iHL is equal to hE.A1;:::;Ad/.�/1; 1iK. Since each solutionof the moment problem is of the form hE.A1;:::;Ad/.�/1; 1iK by Theorem 12.40 andso coincides with � by the preceding considerations, � is the unique representingmeasure of L. Thus � is determinate.

Let M be a Borel set of Rd. By definition CdŒ x � is dense in HL. Hence there is asequence . pn/n2N of polynomials pn 2 CdŒ x � such that pn.x/! E.X1;:::;Xd/

.M/1 inHL L2.�/. Then, by the functional calculus,

jj. pn.x/� E.X1;:::;Xd/.M//1jj2HL

DZRdjpn.�/ � �M.�/j2dhE.X1;:::;Xd/

.�/1; 1iHL ! 0;

so the closure of CdŒ x � in L2.Rd; �/ contains all characteristic functions�M . HenceCdŒ x � is dense in L2.Rd; �/: Thus HL Š L2.Rd; �/, so � is strictly determinate.

It is easily checked that .Xk˙i/CdŒ x � is dense in L2.Rd; �/ if and only if CdŒ x � isdense in L2.Rd; .1Cx2k/d�/. Therefore, by Proposition A.42, since Xk is essentiallyself-adjoint, .Xk ˙ i/CdŒ x � is dense in HL Š L2.Rd; �/. Hence CdŒ x � is dense inL2.Rd; .1C x2k/d�/. This proves that � is strongly determinate.

Conversely, suppose that � is strongly determinate. Then, by definition, for eachk D 1; : : : ; d, CdŒ x � is dense in L2.Rd; .1C x2k/d�/. Hence .Xk ˙ i/CdŒ x � is densein L2.Rd; �/ and so also in its subspace HL. Therefore, Xk is essentially self-adjointon HL again by Proposition A.42. utExample 14.4 (A strictly determinate measure that is not strongly determinate) Let�0 2MC.R/ be an N-extremal solution of an indeterminate moment problem. Wedefine a measure � 2 MC.R/ by d� WD .1 C x2/�1d�0. Since CŒx� is dense inL2.R; �0/ D L2.R; .1 C x2/d�/, � is determinate by Corollary 6.11. Let � be theimage of � under the mapping x 7! .x; x2/ of R into R2. Clearly, � 2MC.R2/.

We prove that � is not strongly determinate. Assume the contrary. Then CŒx1; x2�is dense in L2.R2; .1Cx22/d�/. Then, by the construction of�, CŒx� would be densein L2.R; .1Cx4/d�/ D L2.R; .1Cx4/.1Cx2/�1d�0/. It follows from the inequality.1C x4/.1C x2/�1 � 1

2.1C x2/ that CŒx� is dense in L2.R; .1C x2/d�0/. Therefore

�0 is determinate by Corollary 6.11, which contradicts the choice of �0.

14.2 Polynomial Approximation 361

Now we show that� is determinate. Let Q� 2M�. By construction� is supportedon the graph of the parabola x2 D x21. Since

R.x2 � x21/

2d� D R.x2 � x21/

2d Q� D 0,Q� is also supported on the graph of this parabola (by Proposition 1.23). Let �1. Q�/denote the projection of Q� onto the x1-axis. Since Q� 2 M�, �1. Q�/ has the samemoments as �1.�/ D �. Because � is determinate, we obtain �1. Q�/ D � D �1.�/.This implies that Q� D �, which proves that � is determinate.

Since �0 is N-extremal, CŒx� is dense in L2.R; �0/ D L2.R; .1 C x2/d�/.Therefore, CŒx1; x2� is dense in L2.R2; .1 C x21/d�/ and so in L2.R2; �/. Thus �is strictly determinate.

It is instructive to look at this example from the operator-theoretic perspective.Corollary 6.11 implies that X is essentially self-adjoint and that X2 is not essentiallyself-adjoint on L2.R; �/. Using the definition of � we verify that the multiplicationoperators X1 and X2 by the variables x1 and x2, respectively, on L2.R2; �/ areunitarily equivalent to the operators X and X2 on L2.R; �/, respectively. Hence X2is not essentially self-adjoint, so � is not strongly determinate by Theorem 14.2. ı

We close this section by proving a sufficient criterion for ultradeterminacy.

Proposition 14.5 If � 2 MC.Rd/ and CdŒ x � is dense in Lp.Rd; �/ for somep > 2, then � is ultradeterminate and CdŒ x � is dense in L2.Rd; .1C kxkq/d�/ forany q � 0.Proof Let q � 0, ' 2 Cc.R

d/ and f 2 CdŒ x �. Note that . pp�2 /

�1 C . p2/�1 D 1. We

apply the Hölder inequality and obtain

Zj'�f j2.1C kxkq/ d� �

�Zj'�f jpd�

�2=p�Z.1Ckxkq/ p

p�2 d�

� p�2p

:

(14.1)

Since CdŒ x � is dense in Lp.Rd; �/, there is an f 2 CdŒ x � such thatR j' � f jpd�

is arbitrarily small. The last integral in (14.1) is finite by the assumption� 2 MC.Rd/. Therefore, since Cc.R

d/ is dense in L2.Rd; .1 C kxkq/d�/, sois CdŒ x � by (14.1). Setting q D 2 the latter means that � is ultradeterminate byDefinition 14.1. ut

14.2 Polynomial Approximation

In this section, we develop a number of criteria for determinacy that are based ondensity conditions of polynomials.

For a Borel mapping ' W Rd ! Rm and a Borel measure � on Rd, we denote by'.�/ the image of � under the mapping ', that is, '.�/.M/ WD �.'�1.M// for anyBorel set M of Rm. Then the transformation formula

ZRm

f .y/ d'.�/.y/ DZRd

f .'.x// d�.x/ (14.2)


holds for any function f 2 L1.Rm; '.�//.Let �j.x1; : : : ; xd/ D xj be the j-th coordinate mapping of Rd into R. Then the

measure �j.�/ on R is called the j-th marginal measure of �.The following basic result is Petersen’s theorem.

Theorem 14.6 Let � 2MC.Rd/. If all marginal measures �1.�/; : : : ; �d.�/ aredeterminate, then � itself is determinate.

Proof Suppose that � 2 M�. Let �1; : : : ; �d be characteristic functions of Borelsubsets of R and let p1; : : : ; pd be polynomials in one variable. Using the Hölderinequality we derive

jj�1.x1/ : : :�d.xd/� p1.x1/ : : : pd.xd/jjL1.Rd ;�/

� jj.�1.x1/ � p1.x1//�2.x2/ : : : �d.xd/jjL1.Rd ;�/

C jjp1.x1/.�2.x2/� p2.x2//�3.x3/ : : : �d.xd/jjL1.Rd ;�/ C : : :C jjp1.x1/ : : : pd�1.xd�1/.�d.xd/ � pd.xd//jjL1.Rd ;�/

� jj�1.x1/ � p1.x1/jjL2.Rd ;�/ jj�2.x2/ : : : �d.xd/jjL2.Rd ;�/

C jj�2.x2/ � p2.x2/jjL2.Rd ;�/ jjp1.x1/�3.x3/ : : : �d.xd/jjL2.Rd ;�/ C : : :C jj�d.xd/ � pd.xd/jjL2.Rd ;�/ jjp1.x1/ : : : pd�1.xd�1/jjL2.Rd ;�/:

Let j 2 f1; : : : ; dg. Clearly, � Š � implies that �j.�/ Š �j.�/. Therefore, �j.�/ D�j.�/, because �j.�/ is determinate by assumption. Hence

jj�j.xj/�pj.xj/jjL2.Rd ;�/ D jj�j.xj/�pj.xj/jjL2.R;�j.�// (14.3)

D jj�j.xj/�pj.xj/jjL2.R;�j.�//: (14.4)

Since �j.�/ is determinate, the polynomials CŒxj� are dense in L2.R; �j.�// byTheorem 6.10. Therefore, it follows from (14.3) and (14.4) that we can choosep1 such that jj�1.x1/�p1.x1/jjL2.Rd ;�/ becomes arbitrarily small, then p2 such thatjj�2.x2/�p2.x2/jjL2.Rd ;�/ is small and finally pd such that jj�d.xd/�pd.xd/jjL2.Rd ;�/

is small. Then, by the above inequality,

jj�1.x1/ : : : �d.xd/ � p1.x1/ : : : pd.xd/jjL1.Rd;�/

becomes as small as we want. Since the span of such functions �1.x1/ : : : �d.xd/ isdense in L1.Rd; �/, this shows that the polynomials are dense in L1.Rd; �/. Hence� is an extreme point of M� by Proposition 1.21. Thus each element of M� is anextreme point. Therefore, since �; � 2 M�, we have 1

2.� C �/ 2 M� and hence

� D �. This shows that � is determinate. utIn Exercise 14.7 it is shown that the converse of Theorem 14.6 is not true, that is,

there exists a (strongly) determinate measure with indeterminate marginal measures.Now we derive some corollaries of Theorem 14.6 which provide determinacy

criteria in terms of polynomial approximations.

14.2 Polynomial Approximation 363

Let K be a closed subset of Rd. We restate Definition 2.15 in the present case.

Definition 14.7 A measure � 2MC.Rd/ supported on K is called K-determinate,or determinate on K, if � 2M� and supp � K imply that � D �.

Let ' D .'1; : : : ; 'm/ W Rd ! Rm, where '1; : : : ; 'm 2 RdŒ x �, be a polynomialmapping. If � 2MC.Rd/, then '.�/ 2MC.Rm/ and 'k.�/ 2MC.R/ by (14.2).

Corollary 14.8 Let K be a closed subset of Rd and let � 2 MC.Rd/ be suchthat supp� K. Suppose that the map ' W K ! Rm is injective and 'k.�/ isdeterminate for all k D 1; : : : ;m. Then � is K-determinate.

Proof Let � 2 MC.Rd/ be such that � Š � and supp � K. Since ' is apolynomial mapping and � Š �, it follows from (14.2) that '.�/ Š '.�/. Byassumption, all marginal measures �k.'.�// D 'k.�/ are determinate. Therefore,'.�/ is determinate by Theorem 14.6, so that '.�/ D '.�/.

Let N be a Borel subset of K. Set QN D '.N/. Since ' W K ! Rm is injective,'�1. QN/ D N and hence �.N/ D '.�/. QN/ D '.�/. QN/ D �.N/: Thus, � D �: utCorollary 14.9 Let K be a closed subset of Rd. Assume that there are polynomials'1 : : : ; 'm 2 RdŒ x � which are bounded on K and separate the points of K (that is, ifx; x0 2 K and 'k.x/ D 'k.x0/ for all k D 1; : : : ;m, then x D x0.) Then each measure� 2MC.Rd/ satisfying supp� K is K-determinate.

Proof Define a mapping ' D .'1; : : : ; 'm/ W K ! Rm. Then ' is injective, becausethe 'k separate the points of K: Clearly, supp 'k.�/ 'k.K/ . Since 'k is boundedon K, the set 'k.K/ is compact. Hence 'k.�/ is determinate by Proposition 12.17for k D 1; : : : ; d. Thus Corollary 14.8 applies and gives the assertion. ut

For the next corollary we consider polynomial mappings ' j W Rd ! Rmj

and define ' D .'1; : : : ; 'r/ W Rd ! Rm, where j D 1; : : : ; r, mj 2 N,m D m1C � � �Cmr.

Corollary 14.10 Let � and K be as in Corollary 14.8. If ' W K ! Rm isinjective and all measures '1.�/; : : : ; 'r.�/ are strictly determinate, then � is K-determinate.

Proof We replace in the proof of Corollary 14.8 the map �j on a single coordinateby a map � j on a finite set of coordinates. To be more precise, we write y 2 Rm

as y D .y11; : : : ; y1m1 ; y21; : : : ; yrmr/ and define � j W Rd ! Rmj by � j.y/ D.yj1; : : : ; yjmj/. Then we have the following result: If � 2MC.Rm/ and all measures�1.�/; : : : ; �r.�/ are strictly determinate, then � is determinate.

The proof of this statement follows a similar pattern as the proof of Theo-rem 14.6. The polynomials CŒ yj1; : : : ; yjmj � are dense in L2.Rmj ; � j.�//, because� j.�/ is strictly determinate. We do not carry out the details. Using this generaliza-tion instead of Theorem 14.6 we can argue as in the proof of Corollary 14.8. ut

The presence of “sufficiently” many bounded polynomials on the set supp �can be used to prove stronger results than the plain determinacy. As a sample weconsider subsets of cylinders with compact base sets.


Proposition 14.11 Let K be a closed subset ofRd; d � 2, such that K is a subset ofC �R, where C is a compact set of Rd�1. Suppose that � 2MC.Rd/ is supportedon K. If the marginal measure �d.�/ is determinate, then � is ultradeterminate.

Proof We shall write x 2 Rd as x D .y; xd/with y 2 Rd�1 and xd 2 R. Suppose thatf .y/ and g.xd/ are continuous functions with compact supports. Let us abbreviate

M1 D sup fjg.xd/j W xd 2 Rg; M2 D sup fjp.y/j W y 2 Cg; M3 D sup f1C jjyjj W y 2 Cg:

We denote by jj � jj1 the norm of L2.Rd; .1 C jjxjj2/d�/, by jj � jj2 the norm ofL2.Rd; �/ and by jj � jj3 the norm of L2.R; .1 C x2d/d�d.�//. Let p 2 Cd�1Œy� andq 2 CŒxd�. Using the assumption supp � K C � R and formula (14.2) weobtain

jj f .y/g.xd/� p.y/q.xd/jj1 � jj. f .y/� p.y//g.xd/jj1 C jjp.y/.g.xd/ � q.xd//jj1� M1jj f .y/� p.y/jj1 C M2M3jj.g.xd/ � q.xd//.1C x2d/jj2� M1�.R

d/ sup fj f .y/� p.y/j W y 2 Cg C M2M3jjg.xd/� q.xd/jj3:

By Weierstrass’ approximation theorem, p 2 Cd�1Œ y � can be chosen such thatthe supremum of j f .y/�p.y/j over the compact set C is arbitrary small. Since themarginal measure �d.�/ is determinate, CŒxd� is dense in L2.R; .1Cx2d/d�d.�// byTheorem 6.10. Hence we find q 2 CŒxd� such that jjg � qjj3 is sufficiently small.By the preceding inequality we have shown that the function f .y/g.xd/ is in theclosure of CdŒ x � in L2.Rd; .1 C jjxjj2/d�/. Since the span of such functions isobviously dense, CdŒ x � is dense in L2.Rd; .1 C jjxjj2/d�/. This means that � isultradeterminate. ut

14.3 Partial Determinacy and Moment Functionals

The results of this section show how partial determinacy can be used to concludethat positive functionals are moment functionals. First we treat the case d D 2.

Theorem 14.12 Let Q CŒx1; x2� be a set of polynomials such that CŒx1; x2� isthe linear span of polynomials p.x1/q.x1; x2/, where p 2 CŒx1� and q 2 Q. LetL be a positive linear functional on CŒx1; x2�. Suppose that for each q 2 Q thepositive linear functional Lq on CŒx1� defined by Lq. p/ WD L. p.x1/.x22 C 1/qq/ isdeterminate. Then L is a moment functional on CŒx1; x2�:

We state the important special case Q D fxk2 W k 2 N0g of Theorem 14.12separately as the first assertion of the following corollary.

14.3 Partial Determinacy and Moment Functionals 365

Corollary 14.13 Suppose that s D .s.n;k//k;n2N0 is a positive semidefinite 2-sequence such that for each k 2 N0 the 1-sequence

.s.n;2kC2// C s.n;2k//n2N0 (14.5)

is determinate. Then s is a moment sequence.If in addition the sequence .s.0;n//n2N0 is determinate, then s is determinate.

Proof Put Q D fxk2 W k 2 N0g. Since s.n;m/ D Ls.xn1xm2 /, the functional Lxk2

has themoment sequence (14.5). Therefore, by Theorem 14.12, Ls is a moment functional,so s is a moment sequence.

The determinacy of the sequence .s.n;2/Cs.n;0// implies the determinacy of .s.n;0//(see Exercise 6.3). Hence, if .s.0;n// is also determinate, both marginal sequences ofs are determinate, so s is determinate by Theorem 14.6. ut

The proof of Theorem 14.12 is based on two operator-theoretic lemmas. Inthe proof of Lemma 14.14 we use the self-adjoint extension theory of symmetricoperators (see [Sm9, Section 13.2] or [RS2, Section X.1]). If A is a self-adjointoperator, the unitary operatorUA WD .A� iI/.AC iI/�1 is called the Cayley transformof A.

Lemma 14.14 Suppose that Aj; j 2 J; is a family of pairwise strongly commutingself-adjoint operators and B is a densely defined symmetric operator on a HilbertspaceH such that UAjB.UAj/

� D B for j 2 J. Then there exists a family QAj; j 2 J; ofstrongly commuting self-adjoint operators and a self-adjoint operator QB on a Hilbertspace K such that H is a subspace of K, B QB, Aj QAj, and QAj and QB stronglycommute for all j 2 J.

Proof The proof uses the “doubling trick” from operator theory (see e.g. [Sm9,Example 13.5]). Define QAj D Aj ˚ Aj for j 2 J and B0 D B˚ .�B/ on the Hilbertspace K D H˚H. Obviously, the self-adjoint operators QAj; j 2 J; strongly commuteas the operators Aj; j 2 J; do by assumption. Without loss of generality we assumethat the operator B is closed. Then B0 is also closed. Let N˙.T/ D ker.T� ˙ iI/denote the deficiency spaces of a symmetric operator T. From the definition of B0 weobtain N˙.B0/ D N˙.B/˚N�.B/: Hence V.'; / WD . ; '/, where ' 2 NC.B/, 2 N�.B/, defines an isometric linear map V of NC.B0/ onto N�.B0/. Then, by[Sm9, Theorem 13.10]), the restriction QB of the adjoint operator .B0/� to the domainD. QB/ WD D.B0/C .I�V/NC.B0/ is a self-adjoint extension of B0.

From the assumption UAjB.UAj/� D B it follows that UAjB

�.UAj/� D B�. This

in turn implies that UAjN˙.B/ D N˙.B/: From N˙.B0/ D N˙.B/ ˚ N�.B/ andthe definition of V we conclude that the Cayley transform U QAj

D UAj ˚ UAj ofQAj maps .I�V/NC.B0/ onto itself. Since .UAj/

�D.B/ D D.B/ by assumption andhence UAjD.B/ D D.B/, we get U QAj

D. QB/ D D. QB/: Combined with the relation

U QAj.B0/

�.U QAj/� D UAjB

�.UAj/� ˚ .�UAjB

�.UAj/�/ D B� ˚ .�B�/ D .B0/�


the latter implies that U QAjQB.U QAj

/� D QB. This yields U QAjUQB.U QAj

/� D UQB and hence

U QAjUQB D UQBU QAj

. Since the Cayley transforms of QAj and QB commute, so do their

resolvents. Hence the self-adjoint operators QAj and QB strongly commute. utLemma 14.15 Let A and B be closed symmetric linear operators on a Hilbert spaceH and let D D.A/ \ D.B/ be a linear subspace such that AD D, BD D,and AB' D BA' for ' 2 D. Further, let Q be a subset of D such that the spanof vectors p.A/q, where p 2 CŒx� and q 2 Q, is dense in H and a core for B:For q 2 Q we denote by Aq the restriction of A to the invariant linear subspaceDq WD f p.A/.BC iI/q W p 2 CŒx�g and by Hq the closure of Dq in H. Suppose thatfor each q 2 Q the symmetric operator Aq is essentially self-adjoint on the HilbertspaceHq. Then the operator A onH is self-adjoint and we have UAB.UA/

� D B.

Proof Let q 2 Q and z 2 CnR. Since Aq is essentially self-adjoint on Hq, the rangeof Aq C zI is dense in Hq by Proposition A.42. Hence for any p 2 CŒx� there existsa sequence . pn/n2N of polynomials pn 2 CŒx� such that

.Aq C zI/pn.A/.BC iI/q D .AC zI/pn.A/.BC iI/q! p.A/.BC iI/q in Hq:

Since B is a symmetric operator, k.BC iI/'k2 D kB'k2 C k'k2 for any ' 2 D.B/.We use this fact for the second equality and derive

jj.AC zI/Bpn.A/q� Bp.A/qjj2C jj.AC zI/pn.A/q� p.A/qjj2

D jjB..AC zI/pn.A/q � p.A/q/jj2 C jj.AC zI/pn.A/q � p.A/qjj2

D jj.BC iI/..AC zI/pn.A/q� p.A/q/jj2

D jj.AC zI/pn.A/.BC iI/q � p.A/.BC iI/qjj2 ! 0:

Hence .AC zI/Bpn.A/q! Bp.A/q and .AC zI/pn.A/q! p.A/q as n!1. Sincethe span of vectors p.A/q is dense in H, the preceding shows that the range of ACzIis dense in H, so the operator A on H is self-adjoint by Proposition A.42.

Because .AC zI/�1 is bounded, it follows that pn.A/q ! .AC zI/�1p.A/q andBpn.A/q! .ACzI/�1Bp.A/q. Since the operator B is closed, the latter implies thatB.AC zI/�1p.A/q D .AC zI/�1Bp.A/q. By assumption, the span of vectors p.A/qis a core for B. Therefore, we obtain B.AC zI/�1 D .AC zI/�1B for all vectors of the domain D.B/. Setting z D i and z D �i, it follows that both operatorsUA D I � 2i.AC iI/�1 and U�1

A D IC 2i.A� iI/�1 map D.B/ into itself. Therefore,UAD.B/ D D.B/. Let ' 2 D.B/. Then WD U�1

A ' 2 D.B/ and

UAB.UA/�' D UAB D .I � 2i.AC iI/�1/B D B.I � 2i.A � iI/�1/ D BUA D B':

This proves that B UAB.UA/�: If ' 2 D.UAB.UA/

�/, then WD .UA/�' 2

D.B/ and therefore ' D UA 2 D.B/: This shows that D.UAB.UA/�/ D.B/.

Combined with the relation B UAB.UA/� the latter yields B D UAB.UA/

�. ut

14.3 Partial Determinacy and Moment Functionals 367

Proof of Theorem 14.12 Our aim is to apply Lemma 14.15. Let A and B be theclosures of the operators X1 and X2, respectively, on .HL; h�; �iL/. Let q 2 Q. Recallthat .HLq ; h�; �iLq/ denotes the Hilbert space of the moment functional Lq on CŒx1�.From the definitions of the functional Lq and the corresponding scalar products weobtain

h p1.x1/; p2.x1/iLq D L. p1.x1/p2.x1/.x22 C 1/qq/ D h p1.A/.BC i/q; p2.A/.BC i/qiL

for p1; p2 2 CŒx1�. From this equality it follows that the operator X1 with domainCŒx1� on the Hilbert space HLq and the operator Aq with domain Dq on the Hilbertspace Hq (in the notation of Lemma 14.15) are unitarily equivalent. Since Lq isdeterminate by assumption, X1 is essentially self-adjoint by Theorem 6.10 andso is Aq for each q 2 Q. Therefore, by Lemma 14.15, A is self-adjoint andUAB.UA/

� D B.From Lemma 14.14, applied in the case of a single operator A, it follows that A

and B have strongly commuting self-adjoint extensions on a larger Hilbert space.Hence, by Theorem 12.40, L is a moment functional. ut

Now we turn to the case when the dimension d is larger than 2. The followingtheorem is the counterpart of Corollary 14.13.

Theorem 14.16 Let s D .sn/n2Nd0be a positive semidefinite d-sequence, where

d � 3. Suppose that for j D 1; : : : ; d�1, k1; : : : , kj�1; kjC1; : : : ; kd 2 N0 the 1-sequence

.s.k1;:::;kj�1;n;kjC1;:::;kd�1;2kdC2/ C s.k1;:::;kj�1;n;kjC1;:::;kd�1;2kd//n2N0 (14.6)

is determinate. Further, suppose that for all numbers j; l 2 f1; : : : ; d�1g, j < l, allsequences of one of the following two sets of 1-sequences

.s.:::;kj�1;n;kjC1;:::;kl�1;2klC2;klC1;::: / C s.:::;kj�1;n;kjC1;:::;kl�1;2kl;klC1;::: //n2N0 ; (14.7)

.s.:::;kj�1;2kjC2;kjC1;:::;kl�1;n;klC1;::: / C s.:::;kj�1;2kj;kjC1;:::;kl�1;n;klC1;::: //n2N0 ; (14.8)

where k1; : : : ; kd 2 N0, are determinate. Then s is a moment sequence.If in addition the sequence .s.0;:::;0;n//n2N0 is determinate, then s is determinate.

Proof The proof is given by some modifications in the proof of Theorem 14.12.Let Aj, j D 1; : : : ; d, denote the closure of the operator Xj on HLs . First we fixj 2 f1; : : : ; d�1g and apply Lemma 14.15 in the case when A D Aj, B D Ad and

Q D fxk11 � � � xkj�1j�1 xkjC1

jC1 � � � xkdd W k1; : : : ; kj�1kjC1; : : : ; kd 2 N0g: (14.9)

Since all sequences (14.6) are determinate, the assumptions of Lemma 14.15 arefulfilled. Hence the operator Aj is self-adjoint and we have UAjAd.UAj/

� D Ad.


In order to show that the self-adjoint operators A1; : : : ;Ad�1 strongly commutewe apply Lemma 14.15 once more. Let j; l 2 f1; : : : ; d�1g, j < l. First we assumethat the sequences (14.7) are determinate. We set A D Aj, B D Al, and defineQ again by (14.9). Since all sequences (14.7) are determinate, it follows fromTheorem 6.10 that the corresponding operators Aq in Lemma 14.15 are self-adjoint,so Lemma 14.15 applies and yields UAjAl.UAj/

� D Al. Since Al is self-adjoint, weget UAjUAl.UAj/

� D UAl and so UAjUAl D UAlUAj . Hence the resolvents of Aj andAl commute, so that Aj and Al strongly commute. In the case when the sequences(14.8) are determinate we interchange the role of j and l and proceed in a similarmanner.

Thus, A1; : : : ;Ad�1 is a family of strongly commuting self-adjoint operatorssuch that UAjAd.UAj/

� D Ad for j D 1; : : : ; d�1. That is, the assumptions ofLemma 14.14 are satisfied with B D Ad. From Lemma 14.14 and Theorem 12.40we conclude that s is a moment sequence. As in the proof of Corollary 14.5, thedeterminacy assertion follows from Theorem 14.6. utRemark 14.17 In the preceding proof of Theorem 14.16 we have shown thefollowing fact, which will be used in the proof of Theorem 14.20 below: If for somej 2 f1; : : : ; dg all 1-sequences (14.6) are determinate, it follows from Lemma 14.15that the operator Aj D Xj is self-adjoint on the Hilbert space HL. Likewise, it wasshown in the proof of Theorem 14.12 that the operator A D X1 is self-adjointon HL. ı

14.4 The Multivariate Carleman Condition

Recall from Sect. 4.2 that Carleman’s condition (4.2) for a positive semidefinite1-sequence .tn/n2N0 is that

1XnD1

t� 12n

2n D C1: (14.10)

For a d-sequence s D .sn/n2Nd0

the 1-sequences

sŒ1� WD .s.n;0;:::;0///n2N0 ; sŒ2� WD .s.0;n;:::;0//n2N0 ; : : : ; s

Œd� WD .s.0;:::;0;n//n2N0

are called marginal sequences of s. If s is the moment sequence of � 2MC.Rd/,then

sŒ j�n DZRd

xnj d�.x1; : : : ; xd/ DZR

yn d�j.�/.y/; n 2 N0; j D 1; : : : ; d;

that is, sŒ j� is the moment sequence of the j-th marginal measure �j.�/ of �. HencePetersen’s Theorem 14.6 can be rephrased by saying that if all marginal sequencessŒ1�; : : : ; sŒd� of a d-moment sequence s are determinate, then is s itself.

14.4 The Multivariate Carleman Condition 369

Definition 14.18 Let s be a positive semidefinite d-sequence. We shall say that s,and equivalently the functional Ls, satisfy the multivariate Carleman condition if allmarginal sequences sŒ1�; : : : ; sŒd� satisfy Carleman’s condition (14.10), that is, if

1XnD1.sŒ j�2n/

� 12n �

1XnD1

Ls.x2nj /

� 12n D1 for j D 1; : : : ; d: (14.11)

The following theorem of Nussbaum is the main result in this chapter. It showsthe exceptional usefulness of the multivariate Carleman condition, which impliesboth the existence and uniqueness of a solution of the moment problem on Rd!

Theorem 14.19 Each positive semidefinite d-sequence s D .sn/n2Nd0safisfying the

multivariate Carleman condition is a strongly determinate moment sequence.

Theorem 14.19 follows at once from the following more general result.

Theorem 14.20 Suppose that s D .sn/n2Nd0is a positive semidefinite d-sequence

such that the first d�1 marginal sequences sŒ1�; : : : ; sŒd�1� fulfill Carleman’s condi-tion (14.10). Then s is a moment sequence.

If in addition the last marginal sequence sŒd� satisfies Carleman’s condition(14.10) as well, then the moment sequence s is strongly determinate.

A crucial technical step for the proofs of Theorems 14.20 and 14.25 is the nextlemma.

Lemma 14.21 Let L be a positive linear functional on CdŒ x � and let q; f 2 RdŒ x �.Suppose that QL.�/ WD L. f �/ is also a positive functional on CdŒ x �. Set tn WD L.qn/and rn WD QL.qn/ D L. fqn/ for n 2 N0. Then, if the sequence t D .tn/n2N0 satisfiesCarleman’s condition (14.10), so does the sequence r D .rn/n2N0 .

Proof Since L and QL are positive functionals, t and r are positive semidefinitesequences and the Cauchy–Schwarz inequality (2.7) holds for L. Hence, for n 2 N0,

t22nC2 D L.qnqnC2/2 � L.q2n/L.q2nC4/ D t2nt2nC4; (14.12)

r22n D L.q2nf /2 D L.q4n/L. f 2/ D t4nL. f2/: (14.13)

If t2k D 0 for some k 2 N0, then (14.12) implies that t2n D 0 for n � k, so r2n D 0

for n � k by (14.13) and the assertion holds trivially. The assertion is also trivial ifL. f 2/ D 0. Thus we can assume that t2n > 0 for all n 2 N0 and L. f 2/ > 0.

From (14.12) it follows thatt2jC2

t2j� t2jC4

t2jC2for j 2 N0. Thus, for k 2 N, we get

t2kC2t0D

kC1YjD1

t2jt2j�2

��t2kC2t2k

�kC1;


that is, tkC12k � tk2kC2t0, so t2kC22k � t2k2kC2t20 and hence

2kpt2k � 2kC2

pt2kC2 k.2kC2/

pt0:

Setting k D 2n, we obtain

t�1=.4n/4n � t�1=.4nC2/4nC2 t�1=.2n.4nC2//

0 � t�1=.4nC2/4nC2 .1C t0/

�1: (14.14)

Using (14.13) and (14.14) we derive

r�1=.2n/2n � t�1=.4n/4n L. f 2/�1=.4n/ � t�1=.4n/4n .1C L. f 2//�1

� 1

2.t�1=.4n/4n C t�1=.4nC2/

4nC2 /.1C t0/�1.1C L. f 2//�1:

Summing over n 2 N, by the Carleman condition for the sequence .tn/, the right-hand side yieldsC1 and so does the left hand side. ut

Applying Lemma 14.21 with q D xj, j D 1; : : : ; d, yields the following

Corollary 14.22 Let L be a positive linear functional on CdŒ x � and let f 2 RdŒ x �.Suppose that QL.�/ WD L. f �/ is also a positive linear functional on CdŒ x �. If Lsatisfies the multivariate Carleman condition, so does QL.Proof of Theorem 14.20 Assume first that d D 2. Since the .s.n;0/ D L.xn1//n2N0

fulfills Carleman’s condition (14.10) by assumption, so does the sequence

.sn;2.kC1/ C s.n;2k//n2N0 D .L..xkC12 C x2k2 /xn1//n2N0

by Lemma 14.21. Hence this 1-sequence is determinate by Carleman’s Theorem 4.3.Therefore, Corollary 14.13 applies and implies that s is a moment sequence.

In the case d � 3 the proof is similar. All sequences in (14.6), (14.7), and (14.8)are of the form .L. fxnj //n2N0 or .L. fxnl //n2N0 for some polynomial f 2 PRdŒ x �2.

Since sŒ1�; : : : ; sŒd�1� satisfy Carleman’s condition by assumption, these sequencesdo so for j; l D 1; : : : ; d � 1 by Lemma 14.21. Hence they are determinate byTheorem 4.3. Thus, the assumptions of Theorem 14.16 are fulfilled; hence s is amoment sequence.

Now suppose that Carleman’s condition (14.10) holds for sŒ1�; : : : ; sŒd�. Then,arguing as in the preceding paragraph, it follows that for j D 1; : : : ; d all sequences(14.6) are determinate. Hence Aj D Xj is self-adjoint on HL for j D; : : : ; d byRemark 14.17. Therefore, by Theorem 14.2, s is strongly determinate. utRemark 14.23 The preceding proof of Theorems 14.20 and 14.19 was quiteinvolved. We used Theorem 14.16 to prove s is a moment sequence. However,if we know already that s is a moment sequence, the determinacy assertion ofTheorem 14.19 follows easily: Since the multivariate Carleman condition implies

14.4 The Multivariate Carleman Condition 371

Carleman’s condition for the marginal sequences, the latter are determinate byCarleman’s Theorem 4.3 and hence s is determinate by Petersen’s Theorem 14.6. ı

The following corollary is the multidimensional counterpart of Corollary 4.11.

Corollary 14.24 Let � 2 MC.Rd/. Suppose that there exists an " > 0 such that

ZRd

e"kxk d�.x/ < C1: (14.15)

Then � 2MC.Rd/ and � is strongly determinate.

Proof The proof is similar to the proof of Corollary 4.11. Let j 2 f1; : : : ; dg andn 2 N0. Then x2nj e�"jxj j � "�2n.2n/Š for xj 2 R (see (4.16)) and hence

ZRd

x2nj d� DZRd

x2nj e�"jxj je"jxj j d� � "�2n.2n/ŠZRd

e"kxk d� < C1: (14.16)

Let p 2 RdŒ x �. Then p.x/ � c.1Cx2n1 C� � �Cx2nd / on Rd for some c > 0 and n 2 N,so (14.16) implies that p is �-integrable. Thus � 2MC.Rd/:

Let s be the moment sequence of �. By (14.16) there is a constant M > 0 suchthat

sŒ j�2n D Ls.x2nj / D

ZRd

x2nj d� � M2n.2n/Š for n 2 N0:

By Corollary 4.10, this inequality implies that sŒ j� satisfies Carleman’s condition(14.10). Therefore, s and � are strongly determinate by Theorem 14.19. ut

The next theorem shows that Carleman’s condition can be used to localize thesupport of representing measures.

Theorem 14.25 Let s be a real d-sequence and f D f f1; : : : ; fkg a finite subset ofRdŒ x �. Suppose that the corresponding Riesz functional Ls is Q.f/-positive (that is,Ls. p2/ � 0 and Ls. fjp2/ � 0 for all j D 1; : : : ; k and p 2 RdŒ x �) and satisfies themultivariate Carleman condtion. Then s is a determinate moment sequence and itsrepresenting measure is supported on the semi-algebraic set K.f/.

Proof By Theorem 14.20, s is a determinate moment sequence. Let � denote itsunique representing measure. It suffices to show that Ls. fjp2/ � 0 for all p 2 RdŒ x �implies that fj.x/ � 0 on supp�. For simplicity we suppress the index j.

Define a linear functional QL on RdŒ x � by QL.�/ D Ls. f �/. Since Ls. fp2/ � 0 forp 2 RdŒ x � by assumption, QL is a positive functional on RdŒ x �. Therefore, since Lsatisfies the multivariate Carleman condition, so does QL by Corollary 14.22. Thus, byTheorem 14.20, QL is also a determinate moment functional; let � be its representingmeasure. Then

ZRd

x˛ f .x/d�.x/ D Ls. fx˛/ D QL.x˛/ D

ZRd

x˛ d�.x/; ˛ 2 Nd0: (14.17)


Put MC WD fx 2 Rd W f .x/ � 0g and M� WD fx 2 Rd W f .x/ < 0g. We denoteby �˙ the characteristic function of M˙ and define measures �˙; �˙ 2 MC.R/ byd�˙ D �˙d� and d�˙ D ˙fd�˙: Then � D �C C �� and hence

Zx2kl d�C.x/ �

Zx2kl d�.x/; k 2 N0; l D 1; : : : ; d:

Therefore, since the moment sequence s of � satisfies the multivariate Carlemancondition, so does the moment sequence of �C and hence the moment sequence of�C by Corollary 14.22. Consequently, �C is determinate by Theorem 14.20.

Since d�C � d�� D fd�C C fd�� D fd�, it follows from (14.17) that

Zx˛ d�C.x/ D

Zx˛ d��.x/C

Zx˛ d�.x/ D

Zx˛d.�� C �/.x/; ˛ 2 Nd

0:

Hence, since �C is determinate, �C D �� C � . From 0 D �C.M�/ � ��.M�/ � 0we obtain ��.M�/ D 0 and so �� D 0.

The latter implies that f .x/ � 0 on supp�. Indeed, if we had f .x0/ < 0 for somex0 2 supp�, then it would follow that �f .x/ � " > 0 on a ball B around x0 and

0 D ��.B/ DZB.�f .x// d��.x/ D

ZB.�f .x// d�.x/ � "�.B/ > 0;

which is a contradiction. utIn the operator-theoretic approach Carleman’s condition is closely related to the

theory of quasi-analytic vectors. We briefly discuss this connection.Let T be a symmetric linear operator on a Hilbert space and x 2 \1

nD1D.Tn/:

Since the operator T is symmetric, it is easily verified that the real sequence

t D .tn WD hTnx; xi/n2N0

is positive semidefinite and hence a moment sequence by Hamburger’s theorem 3.8.The vector x is called quasi-analytic for T (see e.g. [Sm9, Definition 7.1]) if

1XnD1kTnxk�1=n D C1: (14.18)

Note that t2n D hT2nx; xi D kTnxk2 for n 2 N0. Therefore, the vector x is quasi-analytic for T if and only if the sequence t satisfies Carleman’s condition (14.10).

Now suppose that s is a positive semidefinite d-sequence and let L D Ls be thecorresponding positive linear functional on RdŒ x �. Then we obtain

sŒ j�2n D Ls.x2nj / D hX2nj 1; 1iL D kXn

j 1k2L; j D 1; : : : ; d; n 2 N0:

14.5 Moments of Gaussian Measure and Surface Measure on the Unit Sphere 373

Hence the marginal sequence sŒ j� fulfills Carleman’s condition (14.10) if and onlyif 1 is a quasi-analytic vector for the multiplication operator Xj by the variable xj.

Quasi-analytic vectors of commuting symmetric operators are studied in detailin [Sm9, Section 7.4]. Theorem 7.18 therein gives for d D 2 an operator-theoreticapproach to Theorem 14.19 based on quasi-analytic vectors.

14.5 Moments of Gaussian Measure and Surface Measureon the Unit Sphere

In this short section we interrupt the study of determinacy and compute the momentsof two important measures. These formulas are of interest in themselves.

Let � denote the Gaussian measure on Rd given by

d� D .2�/�d=2 e�kxk2=2 dx;

where dx is the Lebesgue measure of Rd. ObviouslyRRd ekxkd�.x/ <1. Therefore,

Corollary 14.24 implies that � is a strongly determinate measure of MC.Rd/:

Further, let � be the surface measure of the unit sphere Sd�1 of Rd: Recall thats˛.�/ and s˛.�/, where ˛ 2 Nd

0, denote the moments of these measures.Let j 2 f1; : : : ; dg. Since both measures are invariant under the transformation

xj 7! �xj; xi 7! xi for i ¤ j, we have s˛.�/ D s˛.�/ D 0 if one number ˛j is odd.Thus it suffices to determine the moments s2˛.�/ and s2˛.�/ for ˛ 2 Nd

0:

We begin with some preliminaries. Set .2k � 1/ŠŠ WD 1 � 3 � � � .2k � 1/ for k 2 N

and .�1/ŠŠ WD 1. Using the formulasR10

e�t2=2 dt D p�2

andR10

te�t2=2 dt D 1

and integration by parts we easily compute for k 2 N0;

Z 1

0

t2ke�t2=2dt Dr�

2.2k � 1/ŠŠ and

Z 1

0

t2kC1e�t2=2dt D kŠ

2k: (14.19)

Further, for the Gamma function � .z/ WD R10 tz�1e�tdt; < z > 0, the following

formulas (see e.g. [RW, p. 278]) hold for k 2 N0:

��kC 1=2� D p� 2�k .2k � 1/ŠŠ and � .kC 1/ D kŠ : (14.20)

Therefore, comparing (14.19) and (14.20) we calculate

Z 1

0

tne�t2=2dt D 2.n�1/=2 ��.nC 1/=2�; n 2 N0: (14.21)

This formula plays an essential role in the proof of the next proposition.


Proposition 14.26 For ˛ 2 Nd0 we have

s2˛.�/ D .2�/�d=2ZRd

x2˛e�kxk2=2dx D 2j˛j��d=2 ��˛1 C 1=2

� � � �� ˛d C 1=2�;

s2˛.�/ DZSd�1

y2˛d�.y/ D 2 � .˛1 C 1=2/ � � �� .˛d C 1=2/� .j˛j C d=2/

:

In particular, �.Rd/ D 1 and �.Sd�1/ D 2 �d=2

� .d=2/ .

Proof Set Q� D .2�/d=2�. Then, using formula (14.21) for n D 2˛j, we derive

s2˛. Q�/ DZRd

x2˛e�kxk2=2dx DdY

jD1

ZR

x2˛j e�x2j =2dxj D 2ddY

jD1

Z 1

0

x2˛j e�x2j =2dxj

D 2d2.2j˛j�d/=2dY

jD1��˛jC1=2

�D 2j˛jCd=2dY

jD1��˛jC1=2

�: (14.22)

This yields the formula for s2˛.�/ stated in the proposition.On the other hand, setting r D kxk, applying the transformation x D ry, y 2 Sd�1;

and using (14.21) for n D 2j˛j C d � 1, we obtain

s2˛. Q�/ DZRd

x2˛e�kxk2=2dx DZ 1

0

r2j˛jCd�1e�r2=2drZSd�1

y2˛d�.y/

D 2.2j˛jCd�2/=2� ..2j˛j C d/=2/ s2˛.�/ D 2j˛j�1Cd=2� .j˛j C d=2/ s2˛.�/:

Inserting on the left the expression from (14.22) we get the formula for s2˛.�/.The formulas for �.Rd/ and �.Sd�1/ are obtained by letting ˛ D 0 and using

that � .1=2/ D p� . ut

14.6 Disintegration Techniques and Determinacy

The results of this section are based on the following disintegration theorem formeasures. Recall that the image of a measure � by a map p is denoted by p.�/.

Proposition 14.27 Suppose that X and T are closed subsets of Euclidean spacesand � is a finite Radon measure on X. Let p W X ! T be a �-measurable mappingand � WD p.�/. Then there exist a mapping t 7! �t of T into the set of Radonmeasures on X satisfying the following three conditions:

(i) supp �t p�1.t/.(ii) �t. p�1.t// D 1��a:e:

14.6 Disintegration Techniques and Determinacy 375

(iii) For each nonnegative Borel function f on X we have

ZXf .x/ d�.x/ D

ZTd�.t/

�ZXf .x/ d�t.x/

�: (14.23)

Proof This is a special case of [Bou, Proposition 2.7.13, Chapter IX]. utWe retain the assumptions and notations of Proposition 14.27 and begin with the

preparations of Theorem 14.29 below.Let A and B be countably generated complex �-algebras of �-integrable functions

on X and �-integrable functions on T, respectively, which contain the constantfunctions. The involution is always the complex conjugation of functions. Supposethat f . p.x// 2 A for all f 2 B.

Let g 2 A. Since (14.23) holds for f D gg andRfd� is finite, jgj2 is �t-integrable

�-a.e. Because A is a countably generated, there is a common�-null set N0 such thatjgj2 is �t-integrable for all g 2 A and t 2 TnN0: For notational simplicity we assumethat N0 D ; (for instance, we can set g D 0 on N0 for all g 2 A); then jgj2 is�t-integrable on T: Since the unital �-algebra A is spanned by elements jgj2, eachf 2 A is �t-integrable on T and (14.23) holds.

We define linear functionals Lt, t 2 T, and L on A by

Lt. f / WDZXf .x/ d�t.x/ and L. f / WD

ZXf .x/ d�.x/ for f 2 A:

Lemma 14.28 For any f 2 A, the function t 7! Lt. f / is in L2.T; �/:

Proof We freely use the properties (i)–(iii) from Proposition 14.27. (ii) implies thatLt.1/ D 1. By the above definitions and the Cauchy–Schwarz inequality we obtain

ZTd�.t/ jLt. f /j2 D

ZTd�.t/

ˇˇ Z

Xf d�t.x/

ˇˇ2

�ZTd�.t/

�ZXj f j2 d�t.x/

� �ZX1 d�t.x/

�

DZTd�.t/ Lt. f Nf /Lt.1/ D

ZTd�.t/ Lt. f Nf / D L. f Nf / < C1: ut

We assume that the following three conditions are satisfied:

(1) The measure � is determinate on T for B, that is, if �0 is another Radonmeasure on T such that

Rf d� D R f d�0 for all f 2 B, then � D �0.

(2) B is dense in L2.T; �/.(3) For �-almost all t 2 T, the measure �t is determinate on p�1.t/ for A, that is, if

�0t is a Radon measure on the fibre p�1.t/ such that

Rg d�t D

Rg d�0

t for allg 2 A, then �t D �0

t.

The main result of this section is the following fibre theorem for determinacy.


Theorem 14.29 Let A and B be as above and retain the preceding notation. If theassumptions (1)–(3) hold, then the measure � is determinate on X for A.

Proof Suppose that � 0 is a Radon measure on X such that

ZXg d� D

ZXg d�0 for g 2 A: (14.24)

Let � D p.�/, �t and �0 D p.� 0/, �0t, respectively, be the corresponding mea-

sures from the disintegration theorem and Lt and L0t the corresponding functionals

for � and � 0, respectively. For f 2 B we have f . p.x// 2 A and we compute

ZTd�0.t/f .t/ D

ZTd�0.t/f .t/

ZXd�0

t.x/ (14.25)

DZTd�0.t/

ZXf .t/ d�0

t.x/ DZXd�0.t/

ZXf . p.x//d�0

t.x/ (14.26)

DZXf . p.x//d�0.x/ D

ZXf . p.x//d�.x/ D

ZTd�.t/f .t/: (14.27)

Here the equality in (14.25) holds, since �0t. p

�1.t// D 1 �0-a.e. by (ii). Forthe second equality in (14.26) we used that supp�0

t p�1.t/ by (i) and hencet D p.x/ �0

t-a.e. on supp�0t. The first two equalities in (14.27) hold by (14.23)

and (14.24). By assumption (1), � is determinate on T for B. Hence the precedingimplies that � D �0.

Recall that B is countably generated and dense in L2.T; �/ by assumption (2).Hence there exist a subset N of N and functions 'n 2 A, n 2 N, such that the subsetf Q'n W n 2 Ng of B is an orthonormal basis of L2.T; �/.

Fix f 2 A. We compute the Fourier coefficients of the function Lt. f / of L2.T; �/with respect to this orthonormal basis by

ZTd�.t/ Lt. f / Q'n.t/ D

ZTd�.t/ Q'n.t/

ZXf .x/ d�t.x/

DZTd�.t/

ZX'n. p.x//f .x/ d�t.x/ D

ZXd�.x/ 'n. p.x//f .x/ D L. 'n. p.x// f .x//:

Since � D �0, the same reasoning with �t replaced by �0t shows that

ZTd�.t/ L0

t. f / Q'n.t/ D L. 'n. p.x// f .x//:

Therefore, both functions Lt. f / and L0t. f / from L2.T; �/ (by Lemma 14.28) have

the same Fourier developments

Lt. f / DXn2N

L. 'n. p/ f / Q'n.t/; L0t. f / D

Xn2N

L. '. p/ f / Q'n.t/

14.6 Disintegration Techniques and Determinacy 377

in L2.T; �/. Consequently, Lt. f / D L0t. f / �-a-e. on T. That is, there is a �-null

subset Mf of T such thatRX f .x/ d�t.x/ D

RX f .x/ d�

0t.x/ for t 2 TnMf . Since A is

countably generated, there is a �-null subset M of T such that the latter holds for allf 2 A and t 2 TnM. But then we conclude from assumption (3) that �t D �0

t �-a.e.on T. Therefore, since � D �0 as shown above, it follows from the disintegrationformula (14.23) that

RX ' d� D

RX ' d�

0 for all nonnegative functions, hence for allfunctions, ' 2 Cc.XIR/. This implies that � D �0. ut

We now specialize the preceding general theorem to the moment problem anddevelop a “reduction procedure” for proving determinacy.

Suppose that X is a closed subset of Rd: Let p1; : : : ; pm 2 RdŒ x � and define amapping p W X ! T by p.x/ D . p1.x/; : : : ; pm.x//, where T is a closed subset of Rm

such that p.X/ T. Since the polynomials pj are continuous and X is closed, eachfibre p�1.t/ is a closed subset of Rm. In the case X D Rm the fibres p�1.t/; t 2 T,are just the real algebraic varieties

p�1.t/ D fx 2 Rn W p1.x/ D t1; : : : ; pk.x/ D tkg:

Further, we set A WD CdŒ x � � CŒx1; : : : ; xd� and B WD CmŒ t � � CŒt1; : : : ; ; tm�. Notethat then the assumption f . p.x// 2 A for f 2 B is fulfilled.

Suppose that � 2MC.Rd/ and supp � X. Let � and �t be the correspondingmeasures from Proposition 14.27. For f 2 RmŒ t � B it follows from equations(14.25)–(14.27) that

ZTf .t/ d�.t/ D

ZXf . p.x// d�.x/ D L. f . p.x/// < C1:

Therefore, � 2 MC.Rm/. As noted above, there is a �-null set N0 such that allg 2 RdŒ x � A are �t-integrable for t 2 TnN0. Thus, �t 2MC.Rd/ �-a.e. on T.

Assumptions (1) and (2) mean that � is determinate on T (by Definition 14.7)and that CmŒ t � is dense in L2.T; �/. In this case we shall call � strictly determinateon T. Assumption (iii) says that �t is determinate on p�1.t/ �-a.e. on T:

Therefore, in the preceding setup Theorem 14.29 can be restated as follows.

Theorem 14.30 If � is strictly determinate on T and �t is determinate on the fibrep�1.t/ for �-almost all t 2 T, then � is determinate on X.

Clearly, measures with compact support are strictly determinate by Weierstrass’theorem. Combined with this fact Theorem 14.30 yields the following corollaries.

Corollary 14.31 If T is compact and the measure �t is determinate on the fibrep�1.t/ for �-almost all t 2 T, then � is determinate on X.

Corollary 14.32 If � is strictly determinate on T and the fibre p�1.t/ is boundedfor �-almost all t 2 T, then � is determinate on X.

By specifying the sets X;T and the polynomials pj, there are a number ofapplications of Theorem 14.30 and Corollaries 14.31 and 14.32. We mention three


applications and retain the notation and the setup introduced before Theorem 14.30.These results are of interest in themselves and illustrate the power of the fibretheorem.

1. Let p1.x/ D x1; : : : ; pm.x/ D xm, m < d, and let X and T be closed subsets of Rd

and Rm, respectively, such that p.X/ T. Then Theorem 14.30 yields:The measure � is determinate on X if � D p.�/ is strictly determinate on T

for CmŒ t � and the measure �t is determinate on p�1.t/ for �-almost all t 2 T.2. Suppose that p1; : : : ; pm 2 RdŒ x � are polynomials which are bounded on the

closed subset X of Rd. Put

˛j D inf f pj.x/ W x 2 Xg; ˇj D sup f pj.x/ W x 2 Xg; T D Œ˛1; ˇ1� � � � � � Œ˛m; ˇm�:

An immediate consequence of Corollary 14.31 is the following assertion:The measure � is determinate on X if the fibre measures �t are determinate on

p�1.t/ for �-almost all t 2 T.3. Let p.x/ D x21 C � � � C x2d, d � 2, and X D Rn, T WD p.X/ D RC. Then the

fibres are circles and hence compact. That � is determinate on RC means that� is Stieltjes determinate. In this case, � is equal to the Friedrichs solution andhence CŒt� is dense in L2.RC; �/. That is, determinacy on RC is the same asstrict determinacy on RC. Thus Corollary 14.32 gives:

The measure � is determinate if � D p.�/ is determinate on RC, orequivalently, if � is Stieltjes determinate.

14.7 Exercises

1. Let � 2 MC.Rd/ and q 2 CdŒ x �. Suppose that q.x/ ¤ 0 for x 2 supp�:Prove that CdŒ x � is dense in L2.Rd; jqj2d�/ if and only if q.x/CdŒ x � is densein L2.Rd; �/.

2. Let �; � 2MC.Rd/ and suppose that � Š �:a. Let X be a real algebraic set in Rd such that supp�X: Show that supp �

X:b. Suppose that p 2 CdŒ x � is bounded on supp �. Prove that p is also bounded

on supp � and that sup fjp.x/j W x 2 supp �g D sup fjp.x/j W x 2 supp �g:3. Let �1; : : : ; �d 2MC.R/. Show � WD �1 ˝ � � � ˝ �d 2MC.Rd/.4. Let� D �1˝� � �˝�d ,�j 2MC.R/: Prove that the following are equivalent:

(i) Each measure �k; k D 1; : : : ; d, is determinate.(ii) � is determinate.

(iii) � is ultradeterminate.

5. Let � 2MC.Rd/ and let L be the corresponding moment functional on CdŒ x �.Let p WD �1x1 C � � � C �xd, where �j 2 R; and define a symmetric operator Ap

14.7 Exercises 379

on HL by Apf D p � f for f 2 CdŒ x �. Prove that if � is ultradeterminate, thenthe operator Ap is essentially self-adjoint.

6. (A strongly determinate measure that is not ultradeterminate [Sm5])Let �0 2 MC.R/ be indeterminate and N-extremal such that supp � \

.�"; "/ D ; for some " > 0 and define � by d� D .1 C x2/�1d�0. Let �j D'j.�/, where '1.x/ D .x; 0/ and '2.x/ D .0; x/; x 2 R: Put � D �1 C �2.a. Prove that � 2MC.R2/ is strongly determinate.b. Prove that � is not ultradeterminate.

Hint: Show that the operator Ax1Cx2 (Exercise 5) is not essentially self-adjoint.

7. (A strongly determinate measure with indeterminate marginal measures[Sm5]).

Let �1 D P1nD0 anıxn be an indeterminate N-extremal measure on R such

that x0 D 0 and xn ¤ 0 for n 2 N. Put �1 D P1nD1 anı.xn;0/ and �2 DP1

nD1 anı.0;xn/.

a. Show that �1; �2 2MC.R2/ are determinate.b. Show that � D �1 C �2 is strongly determinate.c. Show that �1.�/ and �1.�/ are indeterminate.

Hint for a.: Use Exercise 7.13.

8. (A characterization of ultradeterminacy [Fu])Let � 2 MC.Rd/ and let L be its moment functional. Consider the joint

graph GXWDf. f ;X1 f ; : : : ;Xd f / W f 2 CdŒ x �g of operators Xj on HL. Prove that� is ultradeterminate if and only if there are self-adjoint operators A1; : : : ;Ad

on HL such that GX is dense in joint graph GAWDf. f ;A1 f ; : : : ;Ad f / W f 2\d

jD1D.Aj/g: In this case we have Aj D Xj for j D 1; : : : ; d:9. Let �; � 2 MC.Rd/: Suppose that d�Dj f .x/jd� for some bounded Borel

function f on Rd. Show that if � is strictly determinate (strongly determinate,ultradeterminate), so is �:

10. Let f be a polynomially bounded Borel function on Rd, that is, j f .x/j � p.x/on Rd for some p 2 RdŒ x �, and let " > 0: Define �1; �2 2 MC.Rd/ by

d�1 D j f .x/je�"kxkdx and d�2 D j f .x/je�"kxk2dx;

where dx stands for the Lebesgue measure on Rd. Show that �1 and �2 arestrongly determinate measures of MC.Rd/.

11. (Radon measures with bounded density on a semi-algebraic sets K.f/)Prove that the assertions of Corollary 12.29 remain valid if the assumption

“K.f/ is compact” is replaced by the assumption “L� satisfies the multivariateCarleman condition (14.11)”.Hint: Show that L WD cL� � L� also satisfies (14.11) and use Theorem 14.25.


12. Let ' 2 C10 .R

d/: Define a linear functional L on RdŒ x � by

L. p/ D h p.�i@1; : : : ;�i@d/'; 'i; p 2 RdŒ x �;

where @j WD @@xj

and h�; �i denotes the scalar product of L2.Rd/ with respect tothe Lebesgue measure. Prove that L is an indeterminate moment functional.Hint: For proving that L is a moment functional, use the Fourier transform; forthe indeterminacy mimic the proof of Example 6.14.

13. (A sufficient determinacy criterion [PVs2, Theorem 2.1])Let Vd D fz D .z1; : : : ; zd/ 2 Cd W z21 C � � � C z2d D �1g. Let L be a moment

functional on RdŒ x �. For n 2 N, we define

n.L/ D min fL. p2/ W p 2 RdŒ x �n; jp.z/j D 1 for z 2 Vd g:

Since n.L/ � nC1.L/ for n 2 N, the limit .L/ WD limn!1 n.L/ exists.

a. Show that L is determinate if .L/ D 0.b. Give an example of a determinate moment functional such that .L/ > 0.

Hints: For a., reduce this to the one-dimensional case and apply Petersen’stheorem 14.6. For b., use Example 14.4.

14.8 Notes

Strong determinacy and ultradeterminacy have been invented by B. Fuglede [Fu],while strict determinacy was introduced in [PSm]. The main part of Theorem 14.2is an important classical result of A. Devinatz [Dv2]. Example 14.4 is taken from[Sm5]. Proposition 14.5 can be found in [BC1] for d D 1 and in [Fu] for generald. L.C. Petersen’s Theorem 14.6 was proved in [Pt]. Corollary 14.13 is anotherclassical result due to G. Eskin [Es]. Theorem 14.12 is due to J. Friedrich [Fr2].

The multivariate Carleman Theorem 14.19 was proved by A.E. Nussbaum [Nu1]using his theory of quasi-analytic vectors. The proof given in the text is takenfrom [PSm]. Another proof based on a localization techniques is given in [Ms3].Extensions of Carleman’s condition and various ramifications are developed in [DJ].Theorem 14.25 is due to J.B. Lasserre [Ls3].

The fibre theorem for determinacy (Theorem 14.29) and the correspondingapplications in Sect. 14.6 are due to the author; they are contained in [PSm]. Thedeterminacy for moment problems on curves was studied in [PSr].

Chapter 15The Complex Moment Problem

In this chapter, we give a digression into the complex moment problem on Cd.In Sect. 15.1, we discuss the equivalence of the complex moment problem on Cd

and the real moment problem on R2d. In Sect. 15.2, we briefly treat the momentproblems for two important �-semigroups (Zd and N0 � Zd). The operator-theoretic approach to the complex moment problem (Theorem 15.6) is developed inSect. 15.3. In Sect. 15.4, we show that each positive functional on CdŒz; z� satisfyingthe complex multivariate Carleman condition is a determinate moment functional(Theorem 15.11). In Sect. 15.5, moment functionals on CŒz; z� are characterized interms of extensions to a larger algebra (Theorem 15.14). Section 15.6 solves thetwo-sided complex moment problem on the complex plane (Theorem 15.15).

15.1 Relations Between Complex and Real MomentProblems

The complex moment problem on Cd is the moment problem for the �-semigroupN2d0 with pointwise addition as semigroup operation and involution defined by

.m; n/� D .n;m/ for m; n 2 Nd0. Recall from Example 2.3.2 that the map

.m; n/ � .m1; : : : ;md; n1; : : : ; nd/ 7! zmzn WD zm11 : : : zmdd z n11 : : : z

mdd

extends by linearity to a �-isomorphism of the semigroup �-algebra CŒN2d0 � on the

polynomial algebra CdŒz; z� WD CŒz1; z1; : : : ; zd; zd� with involution determined by

.zj/� D zj; .zj/

� D zj; j D 1; : : : ; d:

We will write elements of CdŒz; z� as p.z; z/, where z D .z1; : : : ; zd/, z D.z1; : : : ; zd/.


381

382 15 The Complex Moment Problem

Let s D .sm;n/m;n2Nd0

be a complex multisequence. The corresponding Rieszfunctional Ls is the linear functional on CdŒz; z� defined by

Ls.zmzn/ D sm;n; where m; n 2 Nd

0: (15.1)

Then the complex moment problem asks the following:When does there exists a measure � 2 MC.Cd/ such that all p 2 CdŒz; z� are

�-integrable and

sm;n DZCd

zmzn d�.z/ for m; n 2 Nd0; (15.2)

or equivalently,

Ls. p/ DZCd

p.z; z/ d�.z/ for p 2 CdŒz; z�‹ (15.3)

In this case we call s a complex moment sequence and Ls a moment functional.It is known (by Corollary 2.16) and easily verified that a complex moment

sequence s D .sm;n/m;n2Nd0

is positive semidefinite for the �-semigroup N2d0 , that

is,

Xk;m;l;n2Nd

0

skCn;mCl �k;m � l;n � 0

for each finite complex sequence .�k;m/k;m2Nd0:

Further, by Proposition 2.7, s is positive semidefinite for N2d0 if and only if its

Riesz functional Ls is a positive functional on the �-algebra CdŒz; z�. Therefore, eachmoment functional is a positive functional on CdŒz; z�.

We now discuss the relations between the complex moment problem on Cd andthe real moment problem on R2d. For this reason we also consider the polynomial�-algebra C2dŒ x � WD CŒx1; y1; : : : ; xd; yd� with involution given by

.xj/� D xj; . yj/

� D yj; j D 1; : : : ; d:

Let z D .z1; : : : ; zd/ 2 Cd. We write zj D xjCiyj, where yj D Re zj and yj D Im zj,and define a bijection � of Cd onto R2d by

�.z/ � �.z1; : : : ; zd/ D .x1; y1; : : : ; xd; yd/:

There is a �-isomorphism ˚ of the complex �-algebras CdŒz; z� and C2dŒ x � givenby

˚. p/.x1; y1; : : : ; xd; yd/ D p.x1 C iy1; : : : ; xd C iyd; x1 � iy1; : : : ; xd � iyd/(15.4)

15.1 Relations Between Complex and Real Moment Problems 383

for p.z; z/ 2 CdŒz; z�: Their Hermitian parts

CdŒz; z�h WD fp 2 CdŒz; z� W p� D pg ; C2dŒ x �h WD fp 2 C2dŒ x � W p� D pg D R2dŒ x �

are real algebras and the complex �-algebras CdŒz; z� and C2dŒ x � are the complex-ifications (as defined in Sect. 12.1) of the two real algebras CdŒz; z�h and R2dŒ x �,respectively. Therefore, from Lemma 2.17(i) we obtain

XR2dŒ x �

2 DX

C2dŒ x �2 D ˚�XCdŒz; z�

2� D ˚�X.CdŒz; z�h/

2�:

The �-isomorphism˚ and its inverse˚�1 map quadratic modules and preorderingsof one of these real algebras CdŒz; z�h and R2dŒ x �, respectively, onto quadraticmodules and preorderings of the other. Further, the bijection � of Cd and R2d mapsthe character set Cd of the real algebra CdŒz; z�h onto the character set R2d of the realalgebra R2dŒ x �. Therefore, by means of the mappings � and˚ and their inverses allconcepts and results from real algebraic geometry and its applications to the momentproblem carry over from R2d and R2dŒ x � to Cd and CdŒz; z�h. We do not restate thecomplex versions of these results. Our main aim in this chapter is to develop resultsthat are more specific to the complex case.

We will use the correspondence discussed in the preceding paragraph to derivetwo interesting facts regarding the complex moment problem.

Proposition 15.1 There exists a positive semidefinite sequence s D .sm;n/m;n2N0

on the �-semigroup N20, and equivalently, a positive linear functional Ls on the �-

algebra C1Œz; z� � CŒz; z�, such that s is not a complex moment sequence and Ls isnot a moment functional.

Proposition 15.1 follows at once from Proposition 13.5. The following “com-plex” Haviland theorem is an immediate consequence of the “real” Havilandtheorem 1.12 for R2dŒ x �:

Proposition 15.2 A multisequence s D .sm;n/m;n2Nd0is a complex moment

sequence if and only if Ls. p/ � 0 for all p 2 CdŒz; z� satisfying p.z; z/ � 0

for z 2 Cd.

As noted above, the moment problems on R2d and Cd are equivalent by theisomorphism˚ of CdŒz; z� and C2dŒ x �. For d D 1 we state the formulas relating thecorresponding moments. Let L be a linear functional on C1Œz; z� D CŒz; z� and definea linear functional QL on C2Œ x � D CŒx1; y1� and sequences s D .sm;n/, Qs D .Qsm;n/ by

QL. p/ D L.˚�1. p//; p 2 C2dŒ x �;

sm;n D L.zmzn/; Qsm;n D QL.xm1 yn1/; .m; n/ 2 N20:


Some computations yield the following formulas for .m; n/ 2 N20 :

Qsm;n D 2�m�nmX

kD0

nXlD0

m

k

! n

l

!in�2l skCl;mCn�k�l;

sm;n DmX

kD0

nXlD0

m

k

! n

l

!im�k.�i/n�l QskCl;mCn�k�l:

15.2 The Moment Problems for the �-SemigroupsZd andN0 � Zd

The moment problems for the �-semigroups Zd and N0 �Zd are moment problemson the d-torus Td and on the cylinder set R �Td, respectively.

Let us begin with the �-semigroup Zd from Example 2.3.3. The semigroupoperation of Zd is addition and the involution is given by n� D �n. Further, themap n D .n1; : : : ; nd/ 7! zn D zn11 � � � zndd gives a �-isomorphism of the group �-algebra CŒZd� onto the �-algebra of trigonometric polynomials in d variables. Fornotational simplicity we identify CŒZd� with the latter �-algebra, that is,

CŒZd� D CŒz1; z1; : : : ; zd; zd W z1z1 D z1z1 D 1; : : : ; zdzd D zdzd D 1�:

The characters of the complex �-algebraCŒZd�; or equivalently of its Hermitian partCŒZd�h, are the point evaluations at points of the d-torus

Td D fz D .z1; : : : ; zd/ 2 Cd W jz1j D � � � D jzdj D 1 g:

Let us pass to the “real setting”. Set A WD ˚.CŒZd�h/, where ˚ is definedby (15.4). Then A is a real unital algebra with 2d generators x1; y1; : : : ; xd; yd andrelations hj WD x2j C y2j � 1 D 0, j D 1; : : : ; d. That is, A of the form A D R2dŒ x �=I,

where I is the ideal of R2dŒ x � generated by h1; : : : ; hd. The character space OA is thecompact real algebraic set

OA D Z.I/ D �.Td/ D f.x1; y1; : : : ; xd; yd/ 2 R2d W x21 C y21 D 1; : : : ; x2d C y2d D 1g:

The following result says that strictly positive trigonometric polynomials on Td

are always sums of squares.

Proposition 15.3 Let p 2 CŒZd�. If p.z; z/ > 0 for z 2 Td, then p 2PCŒZd�2:

Proof The assumption implies that p 2 CŒZd�h, so ˚. p/ 2 A D R2dŒ x �=I, and˚. p/.x1; y1; : : : ; xd; yd/ > 0 on OA: Hence it follows from Corollary 12.30(ii) that˚. p/ 2P.R2dŒ x �=I/2 DP A2, so that p 2 ˚�1.

PA2/ DPCŒZd�2. ut

15.2 The Moment Problems for the �-Semigroups Zd and N0 � Zd 385

It is natural ask whether or not each nonnegative trigonometric polynomial on Td

is a sum of squares. This is true if and only if d � 2: For d D 1 this is an immediateconsequence of the Fejér–Riesz theorem 11.1, while for d D 2 this result is muchmore subtle and follows (for instance) from [Ms1, Theorem 9.4.5].

The following proposition solves the moment problem for Zd.

Proposition 15.4 Let s D .sn/n2Zd be a complex sequence and Ls its Riesz func-tional on CŒZd� defined by Ls.zn/ D sn; n 2 Zd: The following are equivalent:

(i) s is a positive semidefinite sequence on the �-semigroup Zd, that is,

Xn;m2Zd

sn�m�m �n � 0

for each finite complex sequence .�n/n2Zd :

(ii) Ls is a positive linear functional on the �-algebra CŒZd�.(iii) s is a moment sequence on Zd, that is, there exists a Radon measure � on Td

such that

sn DZTd

z�n d�.z/ for n 2 Zd:

Proof Proposition 2.7 yields (i)$(ii) and Lemma 2.12 gives (iii)!(ii).The main implication (ii)!(iii) will be derived from Proposition 13.31. Indeed,

we define a linear functional QL on A by QL. p/ D Ls.˚. p//, p 2 CŒZd�h. By (ii),Ls is a positive functional on CŒZd�, hence so is QL on A. Because OA is compact,Proposition 13.31 applies and shows that QL is a moment functional on A. SinceA D ˚.CŒZd�h/ and OA D �.Td/, Ls is a moment functional on CŒZd�. ut

Next we turn to the �-semigroup N0 � Zd with coordinatewise addition assemigroup composition and involution .k; n/� D .k;�n/, where k 2 N0, n 2 Zd. Itis straightforward to check that the semigroup �-algebra CŒN0 � Zd� is isomorphicto the tensor product �-algebra CŒx�˝ CŒZd� and the characters of CŒN0 �Zd� areexactly the point evaluations at points of R �Td.

The next proposition provides the solution of the moment problem for N0 �Zd.

Proposition 15.5 Let s D .sk;n/.k;n/2N0�Zd be a complex sequence and let Ls be itsRiesz functional on CŒN0 �Zd� defined by Ls.xkzn/ D sk;n, .k; n/ 2 N0 �Zd. Thenthe following statements are equivalent:

(i) s is a positive semidefinite sequence on the �-semigroup N0 �Zd, that is,

X.k;m/;.l;n/2N0�Zd

skCl;n�m �k;m � l;n � 0

for each finite complex sequence .�k;n/.k;n/2N0�Zd :

(ii) Ls is a positive linear functional on the �-algebra CŒN0 �Zd�.


(iii) s is a moment sequence on N0 � Zd, that is, there exists a Radon measure �on R �Td such that xkz�n is �-integrable and

sk;n DZR�Td

xkz�n d�.x; z/ for .k; n/ 2 N0 �Zd:

Proof The proof is almost verbatim the same as the proof of Proposition 15.4. Theimplication (ii)!(iii) follows again from Proposition 13.31, but now applied to thereal algebra RŒx�˝ A, where A D ˚.CŒZd�h/. ut

15.3 The Operator-Theoretic Approach to the ComplexMoment Problem

The main aim of this section is to derive Theorem 15.6 below, which is thecounterpart of Theorem 12.40 for the complex moment problem. As Theorem 12.40is about extensions of commuting symmetric operators to strongly commutingself-adjoint operators, Theorem 15.6 deals with extensions of commuting formallynormal operators to strongly commuting normal operators. In order to formulateTheorem 15.6 we need further operator-theoretic considerations. All definitions andresults used in the following discussion can be found in [Sm9, Chapters 4 and 5].

Let Z be a densely defined linear operator on a Hilbert space. The operator Z iscalled formally normal if

D.Z/ D.Z�/ and kZfk D kZ�fk for f 2 D.Z�/

and Z is said to be normal if Z is formally normal and D.Z/ D D.Z�/. Note that Zis normal if and only if Z is closed and Z�Z D ZZ�.

For each normal operator Z there exists a unique spectral measure EZ on theBorel �-algebra of C such that

Z DZC

� dEZ.�/:

If Z1 and Z2 are normal operators acting on the same Hilbert space, we say that Z1and Z2 strongly commute if the spectral measures EZ1 and EZ2 commute, that is,EZ1 .M/EZ2 .N/ D EZ2 .N/EZ1 .M/ for all Borel subsets M;N of C. In this case,

Z1Z2' D Z2Z1' for ' 2 D.Z1Z2/ \D.Z2Z1/; (15.5)

but (15.5) does not imply the strong commutativity of Z1 and Z2. A number ofequivalent formulations of strong commutativity are given in [Sm9, Section 5.6].

15.3 The Operator-Theoretic Approach to the Complex Moment Problem 387

Now suppose that N D .N1; : : : ;Nd/ is a fixed d-tuple of strongly commutingnormal operators N1; : : : ;Nd on a Hilbert space K. Then, by the multidimensionalspectral theorem [Sm9, Theorem 5.21], there exists a unique spectral measure EN

on the Borel �-algebra of Cd such that

Nj DZCd�j dEN.�1; : : : ; �d/ for j D 1; : : : ; d:

Further, if M1; : : : ;Md are Borel subsets of C, then we have

EN.M1 � � � � �Md/ D EN1 .M1/ � � �ENd.Md/: (15.6)

We state basic properties of the functional calculus for the d-tuple N, see [Sm9,Theorem 4.16 and formulas (4.32), (5.32)]. For any measurable function f on Cd

there is a normal operator f .N/ D RCd f .�/ dEN.�/ on K with dense domain

D. f .N// D f' 2 K WZCdj f .�/j2 dhEN.�/'; 'i <1g: (15.7)

Let f ; g be measurable functions on Cd. If ' 2 D. f .N// and 2 D.g.N//, then

hf .N/'; g.N/ i DZCd

f .�/g.�/ dhEN.�/'; i: (15.8)

Further, f .N/� D f .N/. For ˛; ˇ 2 C and � 2 D. f .N// \D.g.N// we have

� 2 D..˛f C ˇg/.N// and .˛f C ˇg/.N/� D ˛f .N/� C ˇg.N/�:

If 2 D.g.N// and 2 D.. fg/.N//, then

2 D. f .N/g.N// and f .N/g.N/ D . fg/.N/:

Let p.N;N�/ denote the operator f .N/ obtained for f .z/ D p.z; z/ 2 CdŒz; z�. Inparticular, Nn is the operator for the function zn, n 2 Nd

0. The linear subspace

D1.N/ WD \n2Nd0D.Nn/:

is dense in K. From (15.7) it follows that all operators p.N;N�/ are defined onD1.N/: The properties of the functional calculus stated above imply that the map

p 7! N. p/ WD p.N;N�/dD1.N/

is a �-representation N of the �-algebra CdŒz; z� on the domain D.N/ WD D1.N/.


Fix ' 2 D1.N/ and define a measure �' 2 MC.Cd/ by �'.�/ WD hEN.�/'; 'i.Let p 2 CdŒz; z�. Formula (15.8), applied with f D p; g D 1; ' D , yields

L'. p/ WD hN. p/'; 'i D hp.N;N�/'; 'i DZCd

p.z; z/ d�'.z/: (15.9)

This shows that L'.�/ D hN.�/'; 'i is a moment functional on CdŒz; z� and �' is arepresenting measure of L' , see (15.3).

We now turn to a special case that is crucial for the moment problem. Supposethat � is a Radon measure on Cd: Let Nj denote the multiplication operator by thecoordinate function zj on the Hilbert space G WD L2.Cd; �/, that is,

.Nj'/.z/ WD zj'.z/ for ' 2 D.Nj/ WD f' 2 G W zj � ' 2 Gg:

Then the adjoint N�j is the multiplication operator by zj and kNj'k D kN�

j 'k for' 2 D.Nj/ D D.N�

j /; that is, Nj is a normal operator.For a Borel subset M of C set Mj D fz 2 Cd W zj 2 Mg. Let �Mj denote the

characteristic function of Mj and define ENj.M/' WD �Mj � ' for ' 2 G. Then ENj

is the spectral measure of Nj, that is, Nj' DRCzj dENj.z/' for ' 2 D.Nj/. Since

the spectral measures ENj and ENk act as multiplications by characteristic functions,they commute, that is, Nj and Nk strongly commute for j; k D 1; : : : ; d: For a Borelset M of Cd let EN.M/ be the multiplication operator by the characteristic functionof M. Then EN is the spectral measure of the d-tuple N D .N1; : : : ;Nd/ of stronglycommuting normal operators N1; : : : ;Nd and we have �.�/ D hEN.�/1; 1i.

The preceding facts hold for any measure � 2 MC.Cd/. Now we assume inaddition that all polynomials of CdŒz; z� are in L2.Cd; �/. Then 1 is in the domainD1.N/ and the above considerations apply to the d-tuple N and ' D 1: Therefore,setting �.�/ D hEN.�/1; 1i, it follows from (15.9) that

L. p/ WD hN. p/1; 1i D hp.N;N�/1; 1i DZCd

p.z; z/ d�.z/ (15.10)

for p 2 CdŒz; z�: Therefore, L is a moment functional and �.�/ D hEN.�/1; 1i is arepresenting measure of L.

Hence L is a positive linear functional on the �-algebra CdŒz; z�. Recall that �L

denotes the GNS representation (see Definition 12.39) associated with L. Let usdescribe the representation �L: By (15.10) we have L. pq/ D R

pq d� for p; q 2CdŒz; z�: Comparing this with (12.31) it follows that the scalar product h�; �iL on thedomain DL of �L is just the scalar product of L2.Cd; �/. Thus DL can be identifiedwith the subspace CdŒz; z� of L2.Cd; �/ and �L acts by �L. p/q D p � q for p 2CdŒz; z�, q 2 DL. In particular, we have �L.zj/ Nj for j D 1; : : : ; d.

Now let L be an arbitrary positive linear functional on CdŒz; z�. Put Zj WD �L.zj/.By (12.29) we have h�L.zj/'; i D h'; �L..zj/�/ i for '; 2 DL, so that

�L.zj/ D �L..zj/�/ �L.zj/

�: (15.11)

15.3 The Operator-Theoretic Approach to the Complex Moment Problem 389

Using once again (12.29) and finally (15.11) we derive for arbitrary ' 2 DL,

kZj'k2 D k�L.zj/'k2 D h�L.zj/'; �L.zj/'i D h'; �L..zj/�/�L.zj/'i

D h'; �L.zj zj/'i D h'; �L.zjzj/'i D h�L..zj/�/'; �L..zj/

�/'iD h�L.zj/

�'; �L.zj/�'i D k�L.zj/

�'k2 D k.Zj/�'k2: (15.12)

This proves that the operator Zj D �L.zj/ is formally normal.Obviously, for any j; k D 1; : : : ; d, the operators Zj and Zk commute on the

domain DL, because zj and zk commute in CdŒz; z� and Zj and Zk leave the domainDL invariant. Thus, .Z1; : : : ;Zd/ is a d-tuple of pairwise commuting formally normaloperators on the dense and invariant domain DL of the Hilbert space HL.

The results established above contain the main technical parts for the followingtheorem on the operator-theoretic approach to the complex moment problem.

Theorem 15.6 Let s D .sm;n/m;n2Nd0be a complex multisequence and let Ls be its

Riesz functional defined by (15.1). The following statements are equivalent:

(i) s is a complex moment sequence, or equivalently, Ls is a moment functional onCdŒz; z�:

(ii) There exists a d-tuple N D .N1; : : : ;Nd/ of strongly commuting normaloperators N1; : : : ;Nd on a Hilbert space G and a vector ' 2 D1.N/ suchthat

sm;n D hNm';Nn'i for m; n 2 Nd0: (15.13)

(iii) s is a positive semidefinite sequence for the �-semigroup N2d0 , or equivalently,

Ls is a positive functional on the �-algebra CdŒz; z�, and there exists a d-tupleN D .N1; : : : ;Nd/ of strongly commuting normal operators N1; : : : ;Nd on aHilbert space G such thatHLs is a subspace of G and

�Ls.z1/ � Z1 N1; : : : ; �Ls.zd/ � Zd Nd : (15.14)

If (iii) holds and EN is the spectral measure of the d-tuple N, then

�.�/ WD hEN.�/1; 1i (15.15)

is a representing measure for s and Ls. Each solution of the complex momentproblem for s resp. Ls is obtained in this manner.

Proof(i)!(iii) Since Ls is a moment functional, it is a positive functional. Let � be

a representing measure for Ls. Let Nj denote the multiplication operator by zj onG WD L2.Cd; �/. Then, as shown above, N D .N1; : : : ;Nd/ is a d-tuple of stronglycommuting normal operators on G such that �Ls.zj/ Nj, j D 1; : : : ; d; and �.�/ D


hEN.�/1; 1i. This proves (iii) and the latter shows that each representing measure �of Ls is of the form (15.15).

(iii)!(ii) By (iii), �Ls.zj/ Nj for j D 1; : : : ; d: Hence

�Ls.zn/ D �Ls.z1/

n1 : : : �Ls.zd/nd Nn1

1 : : :Nndd Nn; n 2 Nd

0: (15.16)

Using (15.1), the definition of the scalar product in DLs , and (15.16) we obtain

sm;n D Ls.znzm/ D h�Ls.z

m/1; �Ls.zn/1i D hNm1;Nn1i; m; n 2 Nd

0;

(15.17)

which is statement (ii) with ' D 1.(ii)!(i) Using first (15.13) and then (15.8) we obtain

sm;n D hNm';Nn'i DZ

zmzn dhEN.z/'; 'i; m; n 2 Nd0; (15.18)

which proves that s is a moment sequence.

Finally, suppose that (iii) holds. From (15.17) and (15.18) with ' D 1 it followsthat �.�/ D hEN.�/1; 1i is a representing measure for s. This completes the proof ofTheorem 15.6. utRemark 15.7 In Theorem 15.6(ii), N1; : : : ;Nd are strongly commuting normaloperators and ' 2 D1.N/. Hence Theorem 15.6 remains valid if (15.13) is replacedby sm;n D hNm.N�/n'; 'i for m; n 2 Nd

0: ıWe restate the main part of Theorem 15.6 in the special case d D 1 separately as

Corollary 15.8 A positive linear functional L on C1Œz; z� D CŒz; z� is a momentfunctional if and only if the formally normal operator Z D �L.z/ has a normalextension acting on a possibly larger Hilbert space G, that is, there exists a normaloperator N on a Hilbert space G such thatHL is a subspace of G and Z N.

If this holds and EN denotes the spectral measure of the normal operator N,then �.�/ D hE.�/1; 1i is a representing measure for L. All solutions of the momentproblem for L are of this form.

A by-product of the study of the complex moment problem is the followingoperator-theoretic result, which is of interest in itself.

Corollary 15.9 There exists a formally normal operator on a Hilbert space whichhas no normal extension on a possibly larger Hilbert space.

Proof From Proposition 15.1, there exists a positive functional L on CŒz; z� that isnot a moment functional. By Corollary 15.8, Z D �L.z/ has the desired properties.

utRemark 15.10 A densely defined linear operator T on a Hilbert space H is calledsubnormal if there exists a normal operator N on a Hilbert space G such that H is

15.4 The Complex Carleman Condition 391

a subspace of G and T N. Using this notion Corollary 15.8 says that a positivelinear functional L on CŒz; z� is a moment functional if and only if the (formallynormal) operator Z D �L.z/ is subnormal. ı

15.4 The Complex Carleman Condition

A complex moment sequence s, likewise its Riesz functional Ls, is called determi-nate if s, or equivalently Ls, has a unique representing measure.

The next theorem is a fundamental result on the complex moment problem.

Theorem 15.11 Let s D .sm;n/m;n2Nd0be a positive semidefinite complex sequence

for the �-semigroup N2d0 . If the complex multivariate Carleman condition

1XnD1

Ls.znj z

nj /

� 12n D 1 for j D 1; : : : ; d (15.19)

holds, then s is a determinate complex moment sequence and its Riesz functional Lsis a determinate moment functional on CdŒz; z�:

Proof For notational simplicity we identify the �-algebras CdŒz; z� and C2dŒ x �by the �-isomorphism ˚ from Sect. 15.1. Then the complex moment problemfor Ls becomes a real moment problem on C2dŒ x � and it suffices to prove thecorresponding assertions for this real moment problem. Our aim is to applyTheorem 14.19.

Recall that zj D xj C iyj, where .xj/� D xj and . yj/� D yj, and Zj D �Ls.zj/.Using the basic property (12.33) of the GNS construction we obtain

Ls.x2nj / D h�Ls.x

2nj /1; 1i D h�Ls.xj/

n1; �Ls.xj/n1i D k�Ls.xj/

n 1k2; (15.20)

Ls.znj z

nj / D h�Ls.z

nj z

nj /1; 1i D h�Ls.zj/

n1; �Ls.zj/n1i D k.Zj/n 1k2: (15.21)

Further, since 2xj D zj C zj, we derive

2n k�Ls.xj/n 1k D k.�Ls.zj/C �Ls.zj//

n 1k D k.Zj C .Zj/�/n 1k

D��

nXkD0

n

k

!.Zj/

�k.Zj/n�k 1

�� nX

kD0

n

k

!k.Zj/�k.Zj/n�k 1k

DnX

kD0

n

k

!k.Zj/k.Zj/n�k 1k D

nXkD0

n

k

!k.Zj/n 1k D 2n k.Zj/n 1k:

Here for the third equality we used that the operators Zj and Z�j commute, so the

binomial theorem applies, while the fourth equality follows from the fact that theoperator Zj is formally normal, that is, from Eq. (15.12).


Combining the preceding estimate with (15.20) and (15.21) we conclude that

Ls.x2nj / � Ls.z

nj z

nj /; n 2 N0:

Almost the same reasoning shows that

Ls. y2nj / � Ls.z

nj z

nj /; n 2 N0:

Therefore, assumption (15.19) implies that the functional Ls on C2dŒ x � satisfies themultivariate Carleman condition (14.11) of the real case, so the assertions followfrom Theorem 14.19. ut

We illustrate the power of Theorem 15.11 by deriving two results for the momentproblem on bounded subsets of Cd. For a sequence s D .sm;n/m;n2Nd

0we set

s. j/n WD s.0;:::;0;n;:::;0/;.0;:::;0;n;:::;0/;

where the number n stands at the places j and d C j. Further, for r D .r1; : : : ; rd/,where r1 > 0; : : : ; rd > 0, let Dr denote the closed polydisc

Dr D fz D .z1; : : : ; zd/ 2 Cd W jzjj � rj; j D 1; : : : ; dg:

Corollary 15.12 A sequence s D .sm;n/m;n2Nd0is a complex moment sequence

which has a representing measure supported on the polydisc Dr if and only if sis positive semidefinite for the �-semigroup N2d

0 and there are numbers Mj > 0

such that

s. j/n � Mjr2nj for n 2 N; j D 1; : : : ; r: (15.22)

Proof It is clear that a complex moment sequence is positive semidefinite for N2d0 .

If s has a representing measure � with support contained in Dr , then

s. j/n DZCd

znj znj d�.z/ � r2nj �.C

d/;

which gives (15.22).Now we prove the if direction. From the definition of s. j/n we get s. j/n D Ls.znj z

nj /.

For sufficiently large n we have Mj � 22n and therefore by (15.22),

Ls.znj z

nj /

12n D .s. j/n /

12n � 2rj:

Hence the Carleman condition (15.19) is fulfilled, so s is a complex momentsequence by Theorem 15.11. Let � be a representing measure of s.

15.5 An Extension Theorem for the Complex Moment Problem 393

To show that � is supported on Dr we use the correspondence between momentproblems of Cd and R2d and apply Proposition 12.15 with gj.x/ D x2j C y2j Š zjzjfor j D 1; : : : ; d: Then (15.22) implies (12.15) and the set G defined by (12.14) isjust the polydisc Dr. Hence supp� Dr by Proposition 12.15. utCorollary 15.13 A sequence s D .sm;n/m;n2Nd

0is a complex moment sequence

on Cd which has a representing measure with compact support if and only if sis positive semidefinite for the �-semigroup N2d

0 and there is a constant c > 0 such

that the sequences .cns. j/n /n2N0 , j D 1; : : : ; d, are bounded.Proof The assertion is easily derived from Corollary 15.12. It suffices to note thatcondition (15.22) holds if and only if all sequences .cns. j/n /n2N0 are bounded, wherec WD .maxj r2j /

�1. ut

15.5 An Extension Theorem for the ComplexMoment Problem

This section deals with the complex moment problem for d D 1, that is, with themoment problem for the �-semigroupN2

0 with involution .m; n/� D .n;m/. Clearly,N20 is a �-subsemigroup of the larger �-semigroup

NC D f.m; n/ 2 Z2 W mC n � 0g with involution .m; n/� D .n;m/:

Hence CŒNd0� is a �-subalgebra of the larger �-algebra CŒNC�.

The following result is the Stochel–Szafraniec extension theorem.

Theorem 15.14 A linear functional L on CŒNd0� D CŒz; z� is a moment functional

if and only if L has an extension to a positive linear functional on CŒNC�.

Proof First we describe the semigroup �-algebra CŒNC�. From its definition it isclear that CŒNC� is the complex �-algebra generated by the functions zmzn on Cnf0g,where m; n 2 Z and mC n � 0. If r.z/ denotes the modulus and u.z/ the phase of z,then zmzn D r.z/mCnu.z/m�n. Then, setting k D mC n, it follows that

CŒNC� D Linfr.z/ku.z/2m�k W k 2 N0;m 2 Zg :

The two functions r.z/ and u.z/ itself are not in CŒNC�, but the functions r.z/u.z/ Dz and v.z/ WD u.z/2 D zz�1 are in CŒNC� and they generate the �-algebra CŒNC�.Writing z D x1 C ix2 with x1; x2 2 R, we get

1C v.z/ D 1C x1 C ix2x1 � ix2

D 2 x21 C i x1x2x21 C x22

; 1 � v.z/ D 2 x22 � i x1x2x21 C x22

:


This implies that the complex algebra CŒNC� is generated by the five functions

x1; x2;x21

x21 C x22;

x22x21 C x22

;x1x2

x21 C x22: (15.23)

The Hermitian part CŒNC�h of the complex �-algebra CŒNC� is the real algebragenerated by the functions (15.23). This real algebra is the special case d D 2 of the�-algebra A treated in Sect. 13.7. Thus, if we identify C with R2 in the obvious way,the assertion of Theorem 15.14 follows at once from Theorem 13.40. ut

In terms of sequences Theorem 15.14 can be rephrased in the following manner:A positive semidefinite sequence c D .cm;n/.m;n/2N2

0for the �-semigroup N2

0 is

a moment sequence for N20 if and only if there is a positive semidefinite sequence

Qc D .Qcm;n/.m;n/2NCfor the �-semigroup NC such that Qcm;n D cm;n for .m; n/ 2 N2

0.Using this reformulation we now give a proof of Theorem 15.14 in the context

of �-semigroups.

Second Proof of Theorem 15.14 This proof makes essential use of the �-semigroupN0 � Z with involution .k; l/� D .k;�l/ and its �-subsemigroup S given byS D f.k; l/ 2 N0 �Z W kC l eveng. It is straightforward to check that the map

! W NC ! S; !.m; n/ WD .mC n;m� n/

is a �-isomorphism of NC and S.First we suppose that c has an extension to a positive semidefinite sequence Qc for

NC. We define a function ' W N0 �Z W! R by

'..k; l// WD Qc!�1.k;l/ if .k; l/ 2 S; (15.24)

'..k; l// WD 0 if .k; l/ 2 N0 � Z; .k; l/ … S: (15.25)

We show that ' is a positive semidefinite function on the �-semigroup N0 � Z.Suppose that finitely many elements si 2 N0�Z and complex numbers �i are given.Set u WD .0; 1/. For s; t 2 N0 � Z we easily verify that s C t 2 S if and only ifs; t 2 S or s; t … S. Further, sC u 2 S if s … S. Since the sequence Qc is positivesemidefinite on NC and !�1 is a �-isomorphism of S onto NC, the restriction of 'to S is positive semidefinite on S. Using the preceding facts we derive

Xi;j

'.s�i C sj/ � i �j D

Xs�i Csj2S

'.s�i C sj/ � i �j

DX

si;sj2S'.s�

i C sj/ � i �j CX

si;sj…S'.s�

i C sj/ � i �j

DX

si;sj2S'.s�

i C sj/ � i �j CX

si;sj…S'..siCu/� C .sjCu// � i �j � 0;

that is, ' is positive semidefinite on N0 �Z.

15.6 The Two-Sided Complex Moment Problem 395

Hence, by Proposition 15.5, .'.k; l//.k;l/N0�Z is a moment sequence for N0 � Z.Let � be a representing measure for this sequence and let � denote its image underthe mapping R �T 3 .r; z/ 7! rz 2 C. Using (15.24) and (15.25) we obtain

cm;n D '.!.m; n// DZR�T

rmCnzn�md�.r; z/

DZR�T

.rz/m.rz/nd�.r; z/ DZC

zmznd�.z/; m; n 2 N0;

that is, c is a moment sequence for N20.

Conversely, assume that c is a moment sequence for N20 and let � be a

representing measure. Then, for m; n 2 N0, we have

cm;n D �.f0g/ımCn;0 CZCnf0g

zmznd�.z/: (15.26)

Since � is a representing measure for c, the function zmz n is �-integrable on Cnf0gfor mC n � 0, that is, for .m; n/ 2 NC. Therefore, the right-hand side of (15.26) iswell-defined for .m; n/ 2 NC; let Qcm;n denote the corresponding number. It is easilyverified that the second summand in (15.26) defines a positive semidefinite sequencefor NC. Obviously, the map .m; n/ 7! ımCn;0 defines a character of NC. Hence thefirst summand in (15.26) is a positive semidefinite sequence for NC as well. Thus,Qc D .Qc.m;n// is a positive semidefinite sequence for the �-semigroup NC: ut

15.6 The Two-Sided Complex Moment Problem

The two-sided complex moment problem is the moment problem for the �-semigroup Z2 with involution .m; n/� D .n;m/.

It is easily verified that the map .m; n/ 7! zmzn gives a �-isomorphism of thesemigroup �-algebra CŒZ2� on the �-algebra CŒz; z; z�1; z�1� of complex Laurentpolynomials in z and z. For notational simplicity we identify these �-algebras. Thecharacter space of CŒZ2� consists of evaluation functionals at points of C� WDCnf0g:

Let s D .sm;n/.m;n/2Z2 be a complex sequence. As usual, Ls is the Riesz functionalon CŒZ2� defined by Ls.zmzn/ D sm;n, .m; n/ 2 Z2.

Then the two-sided complex moment problem is the following question:When does there exist a Radon measure � on C� such that the function zmzn on

C� is �-integrable and

smn DZC�

zmzn d�.z/ for .m; n/ 2 Z2;


or equivalently,

Ls. p/ DZC�

p.z; z/ d�.z/ for p 2 CŒZ2� D CŒz; z; z�1; z�1� ‹

Note that this requires conditions for the measure � at infinity and at zero. In theaffirmative case we call s a moment sequence for Z2 and Ls a moment functional.

The following fundamental result is Bisgaard’s theorem.

Theorem 15.15 A linear functional L on CŒZ2� is a moment functional if and onlyif L is a positive functional, that is, L. f �f / � 0 for all f 2 CŒZ2�.

In terms of �-semigroups the main assertion of this theorem says that eachpositive semidefinite sequence on Z2 is a moment sequence on Z2. This result isreally surprising, since C� has dimension 2 and no additional condition (in terms ofpositivity or of some appropriate extension) is required.

Proof The only if part is obvious. We prove the if part.First we describe the semigroup �-algebra CŒZ2� in terms of generators. Clearly,

a vector space basis of CŒZ2� D CŒz; z; z�1; z�1� is the set fzkzl W k; l 2 Zg. Writingz D x1 C ix2 with x1; x2 2 R we get

z�1 D x1 � ix2x21 C x22

and z�1 D x1 C ix2x21 C x22

:

Hence CŒZ2� is the complex unital algebra generated by the four functions

x1; x2; y1 WD x1x21 C x22

; y2 WD x2x21 C x22

(15.27)

on R2nf0g. Note that all four functions are unbounded on R2nf0g.Let A be the Hermitian part of the complex �-algebra CŒZ2�. Then A is a real

algebra and its character set OA is given by the point evaluations �x at x 2 R2nf0g:(Obviously, �x is a character for x 2 R2nf0g: Since . y1 C iy2/.x1 � ix2/ D 1, thereis no character � on A for which �.x1/ D 0 and �.x2/ D 0.)

The three functions

h1.x/ D x1y1 D x21x21 C x22

; h2.x/ D x2y2 D x22x21 C x22

; h3.x/ D x1y2 D x2y1 D x1x2x21 C x22

are elements of A and they are bounded on OA D f�x W x 2 R2; x ¤ 0 g:To apply Theorem 13.10 we consider a nonempty fibre set given by hj.x/ D �j,

where �j 2 R for j D 1; 2; 3: Then �1 C �2 D 1, so we can assume withoutloss of generality that �1 ¤ 0: In the quotient algebra A=I� we then havex1y1 D �1 ¤ 0, so that y1 D �1x�1

1 , and x2y1 D x1y2 D �3, so that y2 D �3x�11

and x2 D �3��11 x1. Thus the algebra A=I� is generated by x1 and x�1

1 and hence

15.7 Exercises 397

a quotient of the algebra RŒx1; x�11 � of Laurent polynomials. Since

PRŒx; x�1�2

obeys (MP) by Theorem 3.16, so does the preorderingP.A=I�/2 of its quotient

algebra A=I� by Corollary 13.16. Therefore,P

A2 satisfies (MP) by Theorem 13.10.By definition, this means that each positive functional on A, hence on CŒZ2�, is amoment functional. utRemark 15.16 The generators x1; x2; y1; y2 of the algebra A satisfy the relations

x1y1 C x2y2 D 1 and .x21 C x22/. y21 C y22/ D 1: ı

15.7 Exercises

1. Let s D .sm;n/m;n2Nd0

be a positive semidefinite sequence for N2d0 . Show that

jskCm;lCnj2 � skCl;kCl smCn;mCn for k; l;m; n 2 Nd0.

2. Let s D .sm;n/m;n2N0 be a positive semidefinite sequence for N20. Show that there

is a measure � 2MC.R/ supported on RC such that sn;n DRxn d� for n 2 N0.

3. Let � 2 MC.Cd/. Suppose that all polynomials p 2 CdŒz; z� are �-integrable.Define � 2 MC.Cd/ by d� D .1Ckzk2/�1d�, where kzk2 WD z1z1C � � � C zdzd.Show that the complex moment sequence of � is determinate.

4. Suppose that � 2 MC.Cd/ satisfiesRe"kzk2 d� < 1. Show that all moments of

� are finite and the moment sequence of � is determinate.5. Let s D .sm;n/m;n2N0 be a complex sequence and let R > r > 0. Give

necessary and sufficient conditions for s to be a complex moment sequence witha representing measure supported on the following set K:

a. K WD fz 2 C W r � jzj � Rg.b. K WD fz 2 C W Im z � 0; r � jzj � Rg.

6. Let s be a positive semidefinite sequence for the �-semigroup N40. Show that s is

a complex moment sequence on C2 with representing measure supported on theclosed ball f.z1; z2/ 2 C W jz1j2 C jz2j2 � r2g if and only if there is a constantM > 0 such that

nXkD0

2n

k

!s2k;4n�2k;2k;4n�2k � Mr4n for n 2 N0:

7. Let s D .sm;n/m;n2N0 be a complex moment sequence with representing measure� supported on D D fz 2 C W jzj � 1g. What can be said about the support of �if

a. s1;1 D s0;0,b. sm;n D sm�n;0 for m � n,c. s1;1 D s2;0,d. sm;n D smCn;0 for m; n 2 N0?


8. Let s be a complex moment sequence and let � be a representing measure of s.For p 2 CŒz; z�we define a Radon measure �p on C by d�p.z/ D j p.z; z/j2d�.z/.Express the moment sequence of �p in terms of p and s.

15.8 Notes

The interplay between the complex moment problem and subnormality has beenknown and investigated for a long time by operator-theorists [Br, Fo]. Theorem 15.6for d D 1 goes back to Y. Kilpi [Ki], see also [StS2, Proposition 3]. Normalextensions of formally normal unbounded operators have been extensively studiedby J. Stochel and F.H. Szafraniec [StS1, StS2, StS3].

The two formulas at the end of Sect. 15.1 are taken from [StS4]. Proposition 15.5is due to A. Devinatz [Dv1]. The existence of formally normal operators withoutnormal extensions (Corollary 15.9) was discovered in [Cd], see [Sm3] and [St1]for simple explicit examples. The complex Carleman condition was investigated in[StS1]; Theorem 15.11 can be found therein. A solution of the moment problem fordiscs without using Carleman’s condition is given in [Sf].

Theorem 15.14 was proved in [StS4]; our second proof is the original proof in[StS4], while our first proof is taken from [Sm11]. Theorem 15.15 is due to T.M.Bisgaard [Bi]; the very short proof in the text is also from [Sm11].

Chapter 16Semidefinite Programming and PolynomialOptimization

“Finding” the minimum or infimum pmin of a real polynomial p over asemi-algebraic set K.f/ is a basic optimization problem. Sum of squaresdecompositions of polynomials by means of Positivstellensätze and momentproblem methods provide powerful tools for polynomial optimization. The aimof this chapter is to give a short digression into these applications by outlining themain ideas.

In Sect. 16.2, we introduce two relaxations pmomn and psos

n for pmin in terms ofHankel matrices and the quadratic moduleQ.f/ and formulate them as a semidefiniteprogram and its dual. In Sects. 16.3 and 16.4, these relaxations are investigated onK.f/ and Rd, respectively. If the quadratic module Q.f/ is Archimedean and henceK.f/ is compact, a simple application of the Archimedean Positivstellensatz showsthat both relaxations converge to the minimum pmin (Theorem 16.6). Section 16.1contains a brief introduction to semidefinite programming.

In this chapter we use some results of positive semidefinite matrices fromAppendix A.3.

16.1 Semidefinite Programming

Let Symn denote the vector space of real symmetric n � n-matrices and h�; �i thescalar product on Symn defined by hA;Bi D TrAB: Recall that A � 0 means thatthe matrix A is positive semidefinite and A � 0 that A is positive definite.

Now we define a semidefinite program and its dual program.Suppose that a vector b 2 Rm and mC 1 matrices A0; : : : ;Am 2 Symn are given.

Then the primal semidefinite program (SDP) is the following:

p� D infy2Rm

˚bTy W A. y/ WD A0 C y1A1 C � � � C ymAm � 0g: (16.1)


399

400 16 Semidefinite Programming and Polynomial Optimization

That is, one minimizes the linear function bTy D PmjD1 bjyj in a vector variable

y D . y1; : : : ; ym/T 2 Rm subject to the linear matrix inequality (LMI) constraint

A. y/ WD A0 C y1A1 C � � � C ymAm � 0: (16.2)

The set of points y 2 Rm satisfying (16.2) is called a spectrahedron. By Propo-sition A.18, the matrix A. y/ is positive semidefinite if and only if all its principalminors are nonnegative. Since each such minor is a polynomial, the set of pointsfor which A. y/ � 0 is described by finitely many polynomial inequalities. Henceeach spectrahedron is a basic closed semi-algebraic set. By (16.2), a spectrahedronis convex. But a convex basic closed semi-algebraic set is not necessarily aspectrahedron.

Let p 2 RmŒy�: Further, let e 2 Rm be such that p.e/ > 0. Then the polynomialp is called hyperbolic with respect to e if for each y 2 Rm the univariate polynomialp.e C ty/ 2 RŒt� has only real zeros. In this case, the closure of the connectedcomponent of the set fy 2 Rm W p. y/ > 0g containing e is called rigidly convex.If A.e/ � 0, it can be shown that p. y/ WD detA. y/ is a hyperbolic polynomial withrespect to e and the corresponding spectrahedron is rigidly convex. Polyhedra andellipsoids are spectrahedra. The set f. y1; y2/ 2 R2 W y41 C y42 � 1g is not rigidlyconvex, hence it is not a spectrahedron. Spectrahedra form an interesting class ofsets, but their study is outside the scope of this book (see e.g. [BTB]).

If all matrices Aj are diagonal, the constraint A. y/ � 0 consists of inequalitiesof linear functions, so (16.1) is a linear program. Conversely, each linear problembecomes a semidefinite program by writing the linear constraints as an LMI with adiagional matrix. Thus, linear programs are special cases of semidefinite programs.

The dual program associated with (16.1) is defined by

p� D supZ2Symn

f�hA0;Zi W Z � 0 and hAj;Zi D bj; j D 1; : : : ;mg: (16.3)

Thus, one maximizes the linear function �hA0;Zi D �TrA0Z in a matrix variableZ 2 Symn subject to the constraints Z � 0 and hAj;Zi D TrAjZ D bj,j D 1; : : : ;m: It can be shown that the dual program (16.3) is also a semidefiniteprogram.

A vector y 2 Rm resp. a matrix Z 2 Symn is called feasible for (16.1) resp.(16.3) if it satisfies the corresponding constraints. If there are no feasible points, weset p� D C1 resp. p� D �1. A program is called feasible if it has a feasible point.

Proposition 16.1 If y is feasible for (16.1) and Z is feasible for (16.3), then

bTy � p� � p� � �hA0;Zi: (16.4)

16.1 Semidefinite Programming 401

Proof Since y is feasible for (16.1) and Z is feasible for (16.3), A. y/ � 0 and Z � 0.Therefore, hA. y/;Zi � 0 by Proposition A.21(i), so using (16.3) we derive

bTy DmXjD1

yjhAj;Zi D hA. y/;Zi � hA0;Zi � �hA0;Zi:

Taking the infimum over all feasible vectors y and the supremum over all feasiblematrices Z we obtain (16.4). ut

In contrast to linear programming, p� is not equal to p� in general (see Exercise15.4). The number p� � p� is called the duality gap. The next proposition showsthat, under the stronger assumption of strict feasibility, the duality gap is zero.

Proposition 16.2

(i) If p� > �1 and (16.1) is strictly feasible (that is, there exists a vector y 2 Rm

such that A. y/ � 0), then p� D p� and the supremum in (16.3) is a maximum.(ii) If p� < C1 and (16.3) is strictly feasible (that is, there exists a matrix Z 2

Symn such that Z � 0 and hAj;Zi D bj for j D 1; : : : ;m), then p� D p� and theinfimum in (16.1) is a minimum.

Proof We carry out the proof of (i); the proof of (ii) is similar.The subset U of positive definite matrices is an open convex cone in the vector

space SymnC1 (in any norm topology). We define matrices Ai 2 SymnC1

QA0 D�p� 00 A0

�and QAj D

��bj 00 Aj

�; j D 1; : : : ;m;

and consider the convex subset

C WD ˚ QA0 C y1 QA1 C � � � C ym QAm W y1; : : : ; ym 2 R�

of SymnC1. Then U \ C D ;. (Indeed, otherwise A0 C y1A1 C � � � C ymAm � 0 andp� �Pj bjyj > 0 which contradicts the definition of p�.) Therefore, by separationof convex sets (Theorem A.26(i)), there are a linear functional L on SymnC1 and areal number a such that L.A/ � a < L.B/ for A 2 C and B 2 U . Clearly, L ¤ 0.Since U is a cone, it follows that a � 0. Hence L � 0 on C and L > 0 on U . SinceL � 0 on C, we have L. QA0/ � 0 and L. QAj/ D 0 for j D 1; : : : ; n:

By Riesz’ theorem for the Hilbert space .SymnC1; h�; �i/, there exists a matrixQZ 2 SymnC1 such that L.�/ D h�; QZi. Since the closure of U is the cone of positivesemidefinite matrices and L > 0 on U , it follows that L.B/ D hB; QZi D h QZ;Bi � 0for all B � 0. Therefore, QZ � 0 by Proposition A.21(ii). We write QZ as

QZ D�z0 zzT Z

�;


where z0 2 R; z 2 Rn;Z 2 Symn. Then L. QA0/ � 0 and L. QAj/ D 0; j D 1; : : : ;m;yield

hA0;Zi C z0p� � 0 and hAj;Zi D z0bj; j D 1; : : : ;m: (16.5)

A crucial step is to prove that z0 ¤ 0. Assume to the contrary that z0 D 0. Thenz D 0, because QZ � 0. Since L ¤ 0, we have QZ ¤ 0, so Z ¤ 0: By z0 D 0, (16.5)implies hA0;Zi � 0 and hAj;Zi D 0, so that hA. y/;Zi � 0 for all y 2 Rm. But by theassumption of strict feasibility there exists a y 2 Rm such that A. y/ � 0. Fix sucha y. Since QZ � 0, we have Z � 0 and hence hA. y/;Zi � 0 by Proposition A.21(i).Thus, hA. y/;Zi D 0, so that A. y/Z D 0 by Proposition A.21(i). Now A. y/ � 0

implies that A. y/ is invertible. Hence Z D 0, which is a contradiction. This provesthat z0 ¤ 0.

From QZ � 0 we get z0 > 0. Upon scaling we can assume that z0 D 1. From(16.5) we then obtain hAj;Zi D bj for j D 1; : : : ;m, so Z is feasible for the dualprogram (16.3), and�hA0;Zi � p�. Therefore, p� � �hA0;Zi � p�. Since p� � p�by Proposition 16.1, we get p� D p� D �hA0;Zi. This shows that the supremum in(16.3) is a maximum. ut

A large number of problems in various mathematical fields can be formulated interms of semidefinite programming, see e.g. [VB]. We give only two examples.

Example 16.3 (Largest eigenvalue of a symmetric matrix) Let �max.B/ denote thelargest eigenvalue of B 2 Symn. Then

�max.B/ D miny2R fy W . yI � B/ � 0g

gives �max.B/ by a semidefinite program. Since this program and its dual are strictlyfeasible (take y 2 R such that . yI � B/ � 0 and Z D I), Proposition 16.2 yields

�max.B/ D maxZ2Symn

fhB;Zi W Z � 0; hI;Zi D 1g:

A semidefinite program for the sum of the j largest eigenvalues of B is developed in[OW]. ıExample 16.4 (Sos representation of a polynomial) By Proposition 13.2, a poly-nomial f .x/ D P

˛ f˛x˛ 2 RdŒx�2n is in

PRdŒx�2n if and only if there exists a

positive semidefinite matrix G such that f .x/ D .xn/TGxn, where xn is given by

(13.2). If we write xn.xn/T D P

˛2N2n A˛x˛ with A˛ 2 Symd.n/, Eq. (13.5) means

that hG;A˛i D TrGA˛ D f˛ for ˛ 2 Nd0; j˛j � 2n. Therefore, f 2PRdŒx�2n if and

only if there exists a matrix G 2 Symd.n/ such that

G � 0 and hQ;A˛i D f˛ for ˛ 2 Nd0; j˛j � 2n: (16.6)

16.2 Lasserre Relaxations of Polynomial Optimization with Constraints 403

Hence f is a sum of squares if and only if the feasibility condition (16.6) of thecorresponding semidefinite program is satisfied. That is, testing whether or not f isa sum of squares means checking the feasibility of a semidefinite program. ı

16.2 Lasserre Relaxations of Polynomial Optimization withConstraints

In this section and the next, f D f f0; : : : ; fkg is a fixed finite subset of RdŒx� andp 2 RdŒx�; p ¤ 0. For simplicity we assume that f0 D 1 and fj ¤ 0 for all j.

Our aim is to minimize the polynomial p over the semi-algebraic set K.f/: Thatis, we want to “compute”

pmin WD inff p.x/ W x 2 K.f/g: (16.7)

Let n 2 N0. We denote by nj the largest integer such that nj � 12.n � deg. fj//,

where j D 0; : : : ; k; and set

Q.f/n W D� kX

jD0fj�j W �j 2

XRdŒx�

2; deg. fj�j/ � n

�

D� kX

jD0fj�j W �j 2

XRdŒx�

2nj

�:

Let Q.f/�n denote the set of linear functionals L on RdŒx�n satisfying L.1/ D 1 andL.g/ � 0 for all g 2 Q.f/n.

Now we define two relaxations of (16.7), called Lasserre relaxations, by

pmomn WD inf

˚L. p/ W L 2 Q.f/�n

�; (16.8)

psosn WD sup

˚� 2 R W p � � 2 Q.f/n

�: (16.9)

Here we set psosn D �1 if there is no � 2 R such that p � � 2 Q.f/n.

Let us motivate these relaxations. Obviously, pmin is the supremum of numbers� 2 R such that p � � � 0 on K.f/. Clearly, each f 2 Q.f/ is nonnegative on K.f/.Replacing p � � � 0 on K.f/ by the stronger requirement p � � 2 Q.f/n gives thenumber psos

n , which is less then or equal to pmin. Further, by definition, pmin is theinfimum of all evaluations of p at points of K.f/. Taking the infimum over the largerset of functionals Q.f/�n yields the number pmom

n , which is also less than or equal topmin:

Next we want to reformulate the two relaxations (16.8) and (16.9) in terms of asemidefinite program and its dual. Let us begin with (16.8).

In order to describe the functionals L of Q.f/�n we introduce the�dCn

n

�variables

y˛ WD L.x˛/, where ˛ 2 Nd0, j˛j � n. If the functional L is given by a Radon


measure �, then y˛ is just the ˛-th moment of �. But in general L 2 Q.f/�n does notimply that L is given by a Radon measure, so y˛ is only a variable here.

Let L be a linear functional on RdŒx�n. Then, by definition, L belongs to Q.f/�n ifand only if L.1/ D 1 and for j D 0; : : : ; k we have

L. fjq2/ � 0 for q 2 RdŒx�nj : (16.10)

Now we reformulate the conditions (16.10) in terms of the variables y˛. Clearly,L.1/ D 1 means that y0 D 1. Let us fix j D 0; : : : ; k and write fj D P

˛ fj;˛x˛. We

denote by Hnj. fjy/ the type�dCnj

nj

� � �dCnjnj

�-matrix with entries

Hnj. fjy/˛;ˇ WDX

�fj;�y˛CˇC� ; where j˛j � nj; jˇj � nj: (16.11)

That is, Hnj. fy/ is a truncation of the localized infinite Hankel matrix H. fjy/whose entries are defined by (12.12). For q D P

˛ a˛x˛ 2 RdŒx�nj repeating the

computation of (12.13) we derive

L. fjq2/ D

X˛;ˇ;�

fj;�a˛aˇL.x˛CˇC� /D

X˛;ˇ;�

fj;˛aˇa�y˛CˇC�DX˛;ˇ

Hnj. fjy/˛;ˇa˛aˇ:

(16.12)

That is, (16.10) holds if and only if the matrix Hnj. fjy/ is positive semidefinite. LetH.f/. y/ denote the block diagonal matrix with diagonal blocks Hn0 . f0y/;Hn1 . f1y/;: : : ;Hnk. fky/. Then, by the preceding, condition (16.10) is satisfied for allj D 0; : : : ; k if and only if the matrix H.f/. y/ is positive semidefinite.

Note that H.f/. y/ is a symmetric N � N-matrix, where N WD PkjD0

�dCnjnj

�. By

(16.11), all matrix entries of H.f/. y/ are linear functions with real coefficients ofthe variables y˛, where ˛ 2 Nd

0, j˛j � n. Hence, inserting the condition y0 D 1,there are (constant!) real symmetric N � N-matrices A˛ such that

H.f/. y/ D A0 CX

˛2Nd0 ;0<j˛j�n

y˛A˛: (16.13)

We write the polynomial p as p.x/ DP˛ p˛x˛ . Then L. p/ D p0 CP˛¤0 p˛y˛ .

By (16.8), pmomn is the infimum of the function L. p/ in the M WD �dCn

n

�� 1 variablesy˛, where 0 < j˛j � n, subject to the condition that the matrix H.f/. y/ given by(16.13) is positive semidefinite.

Summarizing, the relaxation (16.8) leads to the primal semidefinite program

pmomn � p0 D inf

. y˛/2RM

� X0<j˛j�n

p˛y˛ W A0 CX

0<j˛j�n

y˛A˛ � 0�: (16.14)

16.3 Polynomial Optimization with Constraints 405

Our next aim is to show that (16.9) yields the corresponding dual program.Suppose that � 2 R and p � � 2 Q.f/n, that is,

p � � DkX

jD0fj�j; where �j 2

XRdŒx�

2nj : (16.15)

By Proposition 13.2, �j 2 PRdŒx�2nj if and only if there is a positive semidefinite

matrix Z. j/ D .Z. j/˛;ˇ/j˛j;jˇj�nj of type�dCnj

nj

� � �dCnjnj

�such that

�j.x/ DX

j˛j;jˇj�nj

Z. j/˛;ˇx˛Cˇ:

Let Z be the block diagonal matrix of type N � N with blocks Z.0/; : : : ; ; Z.k/.Clearly, Z � 0 if and only if Z. j/ � 0 for j D 0; : : : ; k.

Recall that fj DP˛ fj;˛x˛: From (16.13) it follows that the j-th diagonal block of

the matrix A˛ has the .ˇ; �/-matrix entryP

ı fj;ı , where the summation is over all ısatisfying ˛ D ˇ C � C ı. Hence, equating coefficients in (16.15) yields

p0 � � DkX

jD0fj;0Z. j/00 D TrA0Z D hA0;Zi;

p˛ DkX

jD0

XˇC�CıD˛

fj;ıZ. j/ˇ;� D TrA˛Z D hA˛;Zi; ˛ ¤ 0:

Thus, taking the supremum of � in (16.9) is equivalent to taking the supremum of��p0 D �hA0;Zi subject to the conditions p˛ D hA˛;Zi, ˛ 2 Nd

0, 0 < j˛j � n.By the preceding, the relaxation (16.9) leads to the corresponding dual program

psosn �p0 D sup

Z2SymN

˚� hA0;Zi W Z � 0; p˛ D hA˛;Zi for 0 < j˛j � n�:

(16.16)

16.3 Polynomial Optimization with Constraints

Let us retain the notation from the preceding section. Some simple properties of thenumbers (16.8) and (16.9) are given in the next lemma.

Lemma 16.5

(i) psosn � psos

nC1 and pmomn � pmom

nC1 for n 2 N0.(ii) psos

n � pmomn � pmin for n 2 N0.


Proof

(i) Obviously,Q.f/n is a subspace of Q.f/nC1. Therefore, p�� 2 Q.f/n implies thatp � � 2 Q.f/nC1, so that psos

n � psosnC1. Since the restrictions of functionals from

Q.f/�nC1 belong to Q.f/�n , it follows that pmomn � pmom

nC1 .(ii) Since polynomials of Q.f/ are nonnegative on K.f/, each point evaluation at

x 2 K.f/ is in Q.f/�n . Hence pmomn � p.x/ for all x 2 K.f/. This implies that

pmomn � pmin.

Let L 2 Q.f/�n . If p�� 2 Q.f/n, then L. p��/D L. p/��L.1/DL. p/�� 0,that is, � � L. p/: Taking the supremum over � and the infimum over L we getpsosn � pmom

n . utAs an application of the Archimedean Positivstellensatz we show that for an

Archimedean module Q.f/ both relaxations converge to the minimum of p.

Theorem 16.6 Suppose that the quadratic module Q.f/ is Archimedean. Then theset K.f/ is compact, so p attains its minimum over K.f/, and we have

limn!1 psos

n D limn!1 pmom

n D pmin: (16.17)

Proof By Corollary 12.9, the semi-algebraic set K.f/ is compact. Fix � 2 R suchthat � < pmin. Then p.x/�� > 0 onK.f/ and hence p�� 2 Q.f/ by the ArchimedeanPositivstellensatz (Theorem 12.36(i)), that is, p � � D P

j fj�j for some elements�j 2PRdŒx�2. We choose n 2 N such that n � deg. fj�j/ for all j D 0; : : : ; k. Thenwe have p�� DPj fj�j 2 Q.f/n and hence psos

n � � by the definition (16.9) of psosn .

Since � < pmin was arbitrary and we have psosn � pmom

n � pmin and psosn � psos

nC1 byLemma 16.5, this implies (16.17). ut

The following propositions describe two interesting situations. The first showsthat the supremum in (16.16) is attained if K.f/ has interior points and the secondsays that we have finite convergence if the quadratic module Q.f/ is stable.

Proposition 16.7 Suppose that K.f/ has a nonempty interior. Then psosn D pmom

n forn 2 N. If pmom

n > �1, then the supremum in (16.16) is a maximum.

Proof Clearly, Q.f/n is of the form (13.26). Therefore, since K.f/ has an interiorpoint, it follows from Lemma 13.48 and Proposition 13.46 that Q.f/n is closed inRdŒx�n.

Suppose that � 2 R and � > psosn . Then p � � … Q.f/n by the definition psos

n :

Therefore, because Q.f/n is a closed convex set in RdŒx�n, Theorem A.26(ii) applies,so there exists a linear functional L on RdŒx�n such that L. p� �/ < 0 and L.g/ � 0for all g 2 Q.f/n. Pick x 2 K.f/ and define L".g/ D L.g/C "g.x/ for g 2 RdŒx�nand " > 0. Then, L" � 0 on Q.f/n and L". p � �/ < 0 if " > 0 is sufficiently small.Further, L".1/ � " > 0. Hence, upon replacing L" by L".1/�1L", we can assume thatL".1/ D 1. Then L" 2 Q.f/�n and hence

pmomn � L". p/ D L". p� �/C L".�/ < L".�/ D �:

16.3 Polynomial Optimization with Constraints 407

Thus we have shown that � > psosn implies � > pmom

n . Therefore, since psosn � pmom

nby Lemma 16.5(ii), the equality psos

n D pmomn holds.

Assume now in addition that pmomn > �1: The set K.f/ has an interior point,

so it contains an open ball U. Define L.g/ D RU g.x/dx for g 2 RdŒx�n, where

dx means the Lebesgue integration on Rd. Since U K.f/; we have L 2 Q.f/�n :Let q D P

˛ a˛x˛ 2 RdŒx�nj ; q ¤ 0. Since fj ¤ 0, hence fjq2 ¤ 0, and U is open,

using (16.12) we obtain

X˛;ˇ

Hnj. fjy/˛;ˇa˛aˇ D L. fjq2/ D

ZU. fjq

2/.x/ dx > 0: (16.18)

Thus Hnj. y/ � 0 for each j, so that H.f/. y/ � 0. This means that the semidefiniteprogram (16.14) is strictly feasible. Therefore, by Proposition 16.2(i), its dualprogram (16.16) attains its maximum. utProposition 16.8 Suppose that the quadratic module Q.f/ is stable. Then thereexists an n0 2 N0, depending only on deg. p/, such that psos

n D psosn0

for all n � n0.

Proof We use the characterization of stability given by condition (�) in Sect. 13.8.It says that for any n 2 N0 there exists a number l.n/ 2 N0 such that each q 2 Q.f/ncan be represented as q D P

j fj�j with �j 2 PRdŒx�2 such that deg. fj�j/ � l.n/.Recall that f1 D 1. Set n0 WD max.deg. p/; l.deg. p///.

Suppose that n � n0. First let psosn D �1. Then, by definition, there is no � 2 R

such that p � � 2 Q.f/n. Hence there is no � 2 R such that p � � 2 Q.f/n0 , sothat psos

n0D �1. Now assume that psos

n > �1. Let � be a real number such that� < psos

n . Then p � � 2 Q.f/n by definition. Since deg. p � �/ � deg. p/, it followsfrom the definition of n0 that p� � 2 Q.f/n0 , so that psos

n0� �. The preceding proves

that psosn0� psos

n . It is obvious that psosn � psos

n0for n � n0. Thus, psos

n D psosn0

. utRemark 16.9 Suppose that K.f/ is compact and Q.f/ is the preorder T.f/: Then Q.f/is Archimedean by Proposition 12.22 and has property (MP) by Theorem 12.25.Therefore, if d � 2 and K.f/ has interior points, Proposition 13.50 implies thatQ.f/ is not stable. Thus, in this case, Proposition 16.7 applies, but Proposition 16.8does not. Likewise, if dimK.f/ � 2, it can be shown that the assumptions ofTheorem 16.6 and Proposition 16.8 exclude each other. ıRemark 16.10 It is natural to look for conditions which imply finite convergence forthe limit limn!1 pmom

n D pmin: The flat extension theory developed in Sect. 17.6below yields such a result:

Let m WD maxf1; deg. fj/ W j D 0; : : : ; kg and n > m. If the infimum in (16.8) isattained at L and rankHn�m.L/ D rankHn.L/, then pmom

k D pmin for all k � n:To prove this we take Theorem 17.38 for granted and apply this result with

n replaced by n � m and QL by L. Since L 2 Q.f/�n , the positivity condition inTheorem 17.38(ii) is fulfilled. Therefore, by Theorem 17.38(i), L has an r-atomicrepresenting measure �, where r D rankHn�m.L/: Using that L. p/ D pmom

n


(because L attains the infimum (16.8)) and L.1/ D R1d� D 1 (by L 2 Q.f/�n )

we derive

pmomn � pmin D pmin

Z1d� �

Zp.x/d� D L. p/ D pmom

n ; (16.19)

so that pmomn D pmin and hence pmom

k D pmin for k � n. Further, from (16.19) itfollows easily that each atom of � minimizes the polynomial p over the set K.f/. ı

16.4 Global Optimization

In this section, we specialize the setup of Sect. 16.2 to the case k D 1; f1 D 1. ThenK.f/ D Rd and (16.7) becomes a global optimization problem on Rd:

pmin WD inf f p.x/ W x 2 Rd g: (16.20)

First we rewrite the two relaxations (16.8) and (16.9) in this case. For this wesuppose that n � deg. p/.

Obviously, Q.f/ D PRdŒx�2; so Q.f/n is the set of � 2 P

RdŒx�2 such thatdeg.�/ � n. Since we assumed that n � deg. p/, it follows from Lemma 13.1 thatp � � 2 Q.f/n if and only if p � � 2PRdŒx�2: Thus

psosn D sup

˚� 2 R W p � � 2

XRdŒx�

2�

(16.21)

does not depend on n � deg. p/. Let us denote the number from (16.21) by psos.Recall that b n

2c is the largest integer not greater than n

2: Then Q.f/�n is the set of

linear functionals L on RdŒx�n such that L.1/ D 1 and L.g2/ � 0 for g 2 RdŒx�b n2 c:

Set y˛ D L.x˛/ for j˛j � n: The matrix H.f/. y/ from Sect. 16.2 is the truncatedHankel matrix Hb n

2 c.L/ with entries Hb n2 c.L/˛;ˇ D y˛Cˇ; j˛j; jˇj � b n2c: Clearly, a

linear functional L on RdŒx�n is in Q.f/�n if and only if Hb n2 c.L/ � 0; so that

pmomn D inf

˚L. p/ W Hb n

2 c.L/ � 0; y0 D 1�:

From Proposition 16.7 and Lemma 16.5(ii) we conclude that

psos � psosn D pmom

n � pmin for n � deg. p/;

that is, both semidefinite programs (16.14) and (16.16) yield the same lower boundpsos for the polynomial p.x/ on Rd.

It is obvious that pmin D �1 if deg. p/ is odd. Now assume that 2k WD deg. p/is even. It may happen that this lower bound is trivial, that is, psos D �1. Thenontrivial case psos > �1 holds if the semidefinite program (16.16) for (16.21) isfeasible. This means that p is, upon adding a constant, a sum of squares in RdŒx�.

16.4 Global Optimization 409

The following proposition gives a necessary condition and a sufficient conditionfor psos > �1. Let p2k denote the homogeneous part of p of degree 2k D deg. p/;k 2 N.

Proposition 16.11 If psos > �1, then we have p2k 2PRdŒx�2. Conversely, if

p2k � c.x21 C � � � C x2d/k 2

XRdŒx�

2 (16.22)

for some constant c > 0, then psos > �1.

Proof First suppose that psos > �1. Then there exists a real number � such that. p � �/ 2PRdŒx�2, say p � � DPi q

2i . Comparing the parts of degree 2k > 0 on

both sides it follows that p2k 2PRdŒx�2.To prove the second assertion we assume that (16.22) holds. In this proof we

write f � g if f � g 2PRdŒx�2 and abbreviate� WD x21 C � � � C x2d: Let

p.x/� p2k.x/ DX

j˛j<2ka˛x

˛:

For ˛ 2 Nd0; j˛j < 2k; we write x˛ D xˇx� with jˇj < k and j� j � k: Since

2a˛x˛ C ja˛j."�2x2ˇ C "2x2� / D ja˛j."�1xˇ C .sign a˛/" x

�/2 � 0

for " > 0, we conclude that there are real numbers b˛ such that

p.x/ � p2k.x/� "2Xj˛jDk

b˛x2˛ C

Xj˛j<k

b˛x2˛: (16.23)

The multinomial theorem implies that �k � x2˛ for ˛ 2 Nd0, j˛j D k: Therefore,

c2�k � "2Pj˛jDk b˛x

2˛ for sufficiently small " > 0. Since p2k � c�k by (16.22), itfollows from (16.23) that

p.x/ � c

2�k C

Xj˛j<k

b˛x2˛: (16.24)

If k D 1, the sum in (16.24) is a constant �, so that . p.x/� �/ 2PRdŒx�2.Now assume that k � 2. For a > 0 set q. y/ D ak � 2�.k�1/.a C y/k C yk.

It can be verified that the polynomial q.x/ is nonnegative on RC (in fact, it attainsits minimum 0 at y D a). Hence q 2 PRŒy�2 C y

PRŒy�2 by Proposition 3.2.

Therefore, setting y D �, we get q.�/ 2PRdŒx�2. Expanding .aC�/k yields

q.�/ D .1 � 2�.k�1//.ak C�k/� 2�.k�1/k�1XjD1

k

j

!a j�k�j � 0: (16.25)


Dividing by 1� 2�.k�1/ > 0, estimating�k by (16.25), and using (16.24) we obtain

p.x/ � c

2.2.k�1/ � 1/k�1XjD1

k

j

!aj�k�j C

X1�j˛j<k

b˛x2˛ C b0 � c

2ak: (16.26)

If j˛j < k, then �j˛j � x2˛ by the multinomial theorem. Hence, if a is sufficientlylarge, the first summand in (16.26) is greater than the second sum. Thus, we havep.x/ � b0 � c

2ak. Setting � D b0 � c

2ak; this yields . p.x/� �/ 2PRdŒx�2. ut

We illustrate the preceding with two examples.

Example 16.12 Consider the Motzkin polynomial p D x21x22.x

21 C x22 � 3/C 1, see

(13.6). Then pmin D 0. Further, there is no � 2 R such that . p� �/ 2PRŒx1; x2�2.Hence psos D �1: But p6 D x41x

22C x21x

42 2

PRŒx1; x2�2. This shows the necessary

condition given in Proposition 16.11 is not sufficient. ıExample 16.13 Let pı WD x21x

22.x

21 C x22 � 3/ C 1 C ı.x61 C x62/ for ı � 0. It can

be shown that . pı/min D ı1Cı : If ı > 0, then . pı/6 D x41x

22 C x21x

42 C ı.x61 C x62/

satisfies condition (16.22). Hence, by Proposition 16.11, we have . pı/sos > �1 forall ı > 0.

Note that limı!C0. pı/min D 0, but limı!C0. pı/sos D �1: (Indeed, otherwisethere exist a � 2 R and a positive null sequence .ın/ such that pın � � is inP

RŒx1; x2�2 for all n. SinceP

RŒx1; x2�2 is closed, then p0 � � D limn. pın � �/would be in

PRŒx1; x2�2, which contradicts Example 16.12.) ı

16.5 Exercises

1. Show that the closed unit disc in R2 and the set f.x1; x2/ 2 R2 W x1x2 � 1g arespectrahedra.

2. Find a semidefinite program that is strictly feasible such that its dual program isnot feasible. What can be said about p� for such a program?

3. Consider the problem of minimizing x1 subject to the constraints x1 � 0 andx1x2 � 1, where .x1; x2/ 2 R2.

a. Write this as a semidefinite program.b. Show that p� D 0 and the infimum in (16.1) is not attained.

4. (A semidefinite program with positive finite duality gap [VB])Consider the problem of minimizing x1 subject to the constraint

0@0 x1 0

x1 x2 0

0 0 x1 C 1

1A � 0:

Show that this is a semidefinite program such that p� D 0 and p� D �1.

16.6 Notes 411

5. Let �min.B/ the smallest eigenvalue of B 2 Symn. Describe ��min.B/ by asemidefinite program and determine the dual program. Show that both programsare strictly feasible.

6. Prove by induction on d that x2k1 � � � C x2kd � 1

2.d�1/.k�1/ .x21 C � � � C x2d/

k 2 RdŒx�2

for k 2 N.7. Find an example p 2 PRŒx1; x2�2 such that pmin D psos D 0, but p2k does not

satisfy condition (16.22) in Proposition 16.11.

16.6 Notes

The semidefinite relaxations and corresponding results are due to J.B. Lasserre[Ls1] who first applied Positivstellensätze and moment methods in polynomialoptimization. Proposition 16.11 is taken from Marshall [Ms2] and parts of ourapproach follow [Ms1]. Semidefinite programming is treated in [V] and [BN].

As noted in this chapter’s introduction, we wanted to give only a small glimpseinto the main ideas. There is now an extensive literature on this area. We refer to thebooks [Ms1, Ls2, BTB, AL] and the articles [Pa, Sw3, La2, Nie].

An important result on finite convergence (that is, pmomn D pmin for large n) of the

relaxation (16.8) was obtained by J. Nie [Nie]. He proved finite convergence underthe assumptions that the quadratic module is Archimedean and standard sufficientoptimality conditions (linear independence of gradients, strict complementarity,second order sufficient condition) hold at each global minimizer. Since theseconditions hold generically, Nie’s theorem implies that (16.8) has finite convergencegenerically if the quadratic module is Archimedean.

Part IVThe Multidimensional Truncated

Moment Problem

Chapter 17Multidimensional Truncated Moment Problems:Existence

This chapter and the next two are devoted to the truncated K-moment problem:Given a subset N of Nd

0, a sequence s D .s˛/˛2N, and a closed subset K of Rd,when does a Radon measure � on K exist such that s˛ D

Rx˛ d� for all ˛ 2 N?

In the affirmative case we call s a truncated K-moment sequence and thecorresponding Riesz functional Ls a truncated K-moment functional.

The existence results developed in this chapter provide important theoreticalinsights into the truncated moment problem, but for proving that a given sequenceis a truncated K-moment sequence their usefulness is limited. While the mainresults in previous chapters (for instance, Theorems 10.1, 10.2 and 12.25, 12.36)contain tractable criteria in terms of Hankel matrices, the solvability conditions ofthis chapter are difficult to verify. One reason is that useful descriptions of strictlypositive polynomials up to a fixed degree 2n are missing. If the strict Positivstel-lensatz (Theorem 12.24) is applied, the degrees of the involved polynomials ofthe preordering exceed 2n in general (see e.g. [Ste2] for an elaborated example indimension one).

In Sects. 17.1 and 17.2, we formulate the truncated K-moment problem and itsprojective version and derive basic existence criteria in terms of positivity conditions(Theorems 17.3, 17.9, 17.10, and 17.15).

Hankel matrices are useful tools for the truncated moment problem. We introduceand study them in Sects. 17.3 and 17.4. In Sect. 17.5, we solve the full momentproblem on RdŒx� with finite rank Hankel matrix. The positive semidefiniteness ofthe Hankel matrix is not sufficient for being a truncated moment functional, butcombined with a flatness condition it is. This is the flat extension theorem of Curtoand Fialkov (Theorems 17.35 and 17.36). It is stated and discussed in Sect. 17.6 andproved in Sect. 17.7.

In this chapter, if not stated otherwise, N denotes a nonempty (finite or infinite)subset of Nd

0, A D fx˛ W ˛ 2 Ng is the set of monomials, A D Lin fx˛ W ˛ 2 Ng istheir linear span, and K is a closed subset of Rd. In Section 17.2, K also denotes a


415

416 17 Multidimensional Truncated Moment Problems: Existence

closed subset of the projective space Pd.R/. Further, we assume the following:

There exists an element e 2 A such that e.x/ � 1 for x 2 K: (17.1)

There are at least two important special cases where condition (17.1) is satisfied.First, if 0 2 N, then (17.1) holds with e.x/ D 1 2 A: The second case is when A isthe vector space of homogeneous polynomials of degree 2n and K is a subset of theunit sphere Sd�1; then e.x/ WD .x21 C � � � C x2d/

n 2 A and e.x/ D 1 on K.

17.1 The TruncatedK-Moment Problemand Existence Criteria

First we recall two standard notations. For a real sequence s D .s˛/˛2N theassociated Riesz functional is the real-valued linear functional Ls on A given byL.x˛/ D s˛ , ˛ 2 N: For a measure � 2 MC.Rd/ such that A L1.Rd; �/, L�

denotes the linear funtional on A defined by

L�. p/ DZRd

p.x/ d�.x/; p 2 A: (17.2)

Definition 17.1 A real sequence sD.s˛/˛2N is called a truncated K-momentsequence if there exists a measure � 2 MC.Rd/ supported on K such that

x˛ 2 L1.Rd; �/ and s˛ DZRd

x˛ d�.x/ for ˛ 2 N: (17.3)

A real-valued linear functional L on A is a truncated K-moment functional if thereis a measure � 2 MC.Rd/ supported on K such that L D L�, that is,

p 2 L1.Rd; �/ and L. p/ DZRd

p.x/ d�.x/ for p 2 A: (17.4)

Each such measure � is called a representing measure of L; the set of representingmeasures is denoted by ML;K.

For K D Rd we call a truncated K-moment sequence simply a truncated momentsequence and a truncated K-moment functional a truncated moment functional, andwe denote the set of representing measures by ML.

Obviously, s satisfies (17.3) if and only if Ls does (17.4). That is, s is a truncatedK-moment sequence if and only if Ls is a truncated K-moment functional. We oftenprefer to work with functionals rather than sequences.

17.1 The Truncated K-Moment Problem and Existence 417

Note that each measure � satisfying (17.2), (17.3), or (17.4) is finite. Indeed,using the element e 2 A from condition (17.1) we obtain

�.Rd/ DZ1 d� �

Ze d� D L.e/ <1:

Thus, the truncated K-moment problem asks: when is a sequence s D .s˛/˛2N atruncated K-moment sequence, or equivalently, when is a linear functional L on A atruncatedK-moment functional? Equations (17.3) and (17.4) are equivalent versionsof the truncated K-moment problem.

In the special case N D Nd0;A D RdŒx� we obtain the (full) multidimensional

K-moment problem. But here our main emphasis will be on finite sets N.Let us relate the truncatedK-moment problem to the moment problem on a finite-

dimensional space E treated in Sect. 1.2. We set X D K and consider the linearsubspace E WD AdK of C.KIR/ spanned by the restrictions of functions f dK, f 2 A.That is, E is the quotient vector space of A by the equivalence relation “f � g if andonly if f .x/ D g.x/ for x 2 K”.

Obviously, if QL is a linear functional on E, then L. f / WD QL. f dK/; f 2 A, definesa linear functional L on A such that

L. f / D 0 for f 2 A; f dK D 0: (17.5)

Conversely, if L is a linear functional on A satisfying (17.5), then there is a well-defined (!) linear functional QL on E given by

QL. f dK/ WD L. f /; f 2 A; (17.6)

and it is clear that L is a truncated K-moment functional according to Definition 17.1if and only if QL is a moment functional on E by Definition 1.1. Further, it is obviousthat L and QL have the same representing measures. This correspondence can beused to restate results from Sect. 1.2 in the present setting. However, one has to becareful: While f 2 A is only zero if f is the null polynomial, g 2 E is zero if andonly if g.x/ D 0 for all x 2 X : This occasionally leads to slight modifications in theformulations of results.

There are two important cases where (17.5) is satisfied. First, if f dK D 0 impliesf D 0; this happens if K has a nonempty interior in Rd. Secondly, by Lemma 17.4,if condition (17.8) holds, in particular, if L is a truncated K-moment functional.

Theorem 17.2 If the set N is finite, then each truncated K-moment functional L onA has a k-atomic representing measure � 2ML;K, where k � jNj D dimA and allatoms of � are in K.

Proof Since L is a truncated K-moment functional, (17.5) is satisfied. Hence theassertion follows from Corollary 1.25, applied to QL and E. ut


The following standard notation is often used in the sequel:

Pos.A;K/ WD f p 2 A W p.x/ � 0 for x 2 Kg: (17.7)

Theorem 17.3 Suppose that K is a compact subset of Rd. A linear functional L onA is a truncated K-moment functional if and only if

L. p/ � 0 for all p 2 Pos.A;K/: (17.8)

In this case the set ML;K of representing measures is vaguely compact.

The following simple fact is used several times.

Lemma 17.4 If a functional L on A satisfies (17.8) (for instance, if L is a truncatedK-moment functional), then (17.5) holds, so the functional QL on E given by (17.6) iswell-defined.

Proof Let f 2 A and f dK D 0. Then ˙f 2 Pos.A;K/ and hence L.˙f / � 0 by(17.8), so that L. f / D 0. This means that (17.5) is satisfied. utProof of Theorem 17.3. The necessity of condition (17.8) is obvious. Conversely,suppose that (17.8) holds. Then the functional QL is well-defined by Lemma 17.4.Since X WD K is compact and condition (17.1) is assumed, it follows fromProposition 1.9, that QL on E, hence L on A, has a representing measure with supportin X D K.

We prove that the set ML;K DMQL is vaguely compact. Let f 2 Cc.KIR/. Thenthere is a constant Mf such that jf .x/j � Mf e.x/ for x 2 K and therefore

sup�2ML;K

ˇ ZKf d�

ˇ � Mf sup�2ML;K

ZKe d� D MfL.e/ D Mf QL.edX / <1:

Hence, MQL DML;K is relatively vaguely compact by Theorem A.6.Let .�i/i2I be a net of measures �i 2ML;K converging vaguely to � 2 MC.K/.

Since K is compact, AdK Cc.KIR/ and hence

ZKf d� D lim

i

ZKf d�i D lim

iL. f / D L. f / for f 2 A;

so that � 2 ML;K. This shows that ML;K is vaguely closed. Being relativelyvaguely compact and vaguely closed, ML;K is vaguely compact. ut

If the set K is only closed and not compact, A is not necessarily an adaptedsubspace of C.K;R/ and in contrast to Haviland’s theorem 1.12 condition (17.8) isnot sufficient for being a truncated K-moment functional. A simple counterexamplein dimension one was given in Example 9.30. Counterparts of Theorem 17.3 forclosed sets K will be obtained in the next section (Theorems 17.13 and 17.15).

17.1 The Truncated K-Moment Problem and Existence 419

For m 2 N we abbreviate

Nd0;m WD f˛ D .˛1; : : : ; ˛d/ 2 Nd

0 W j˛j WD ˛1 C � � � C ˛d � m g: (17.9)

Recall that RdŒx�m are the polynomials p 2 RŒx1; : : : ; xd� such that deg. p/ � m.Our main guiding example for the preceding setup is the following. The reader

might always think of this important special case. Let n 2 N and set N WD Nd0;2n.

Then A D RdŒx�2n and Definition 17.1 has the following form.

Definition 17.5 A sequence s D .s˛/˛2Nd0;2n

is a truncated K-moment sequence if

there is a measure � 2 MC.Rd/ supported on K such that x˛ is �-integrable and

s˛ DZRd

x˛ d�.x/ for ˛ 2 Nd0;2n:

A linear functional L on RdŒx�2n is a truncated K-moment functional if there existsa measure � 2 MC.Rd/ with support in K such that p is �-integrable and

L. p/ DZRd

p.x/ d�.x/ for p 2 RdŒx�2n:

The next theorem restates Theorems 17.2 and 17.3 in this special case. Put

Pos.K/2n WD f p 2 RdŒx�2n W p.x/ � 0 for x 2 Kg: (17.10)

Theorem 17.6 Suppose that K is a compact subset of Rd. A linear functional L onRdŒx�2n is a truncated K-moment functional if and only if

L. p/ � 0 for all p 2 Pos.K/2n: (17.11)

In this case, ML;K is a vaguely compact subset of MC.K/ and L has a k-atomicrepresenting measure, where k � �2nCd

d

�and all atoms of � are in K.

The next result is Stochel’s theorem. It says that solving the truncatedK-momentproblem on RdŒx�2n for all n 2 N leads to a solution of the full K-moment problem.

Theorem 17.7 Let K be a closed subset of Rd and L a linear functional on RdŒx�.Suppose that for each n 2 N the restriction Ln WD LdRdŒx�2n is a truncatedK-moment functional on RdŒx�2n, that is, there exists a measure �n 2 MC.Rd/

supported on K such that

Ln. p/ � L. p/ DZ

p.x/ d�n.x/ for p 2 RdŒx�2n: (17.12)

Then L is aK-moment functional. Further, L has a representing measure� whichis the limit of a subsequence .�nk /k2N in the vague convergence of MC.K/:


Proof Apply Theorem 1.20 to X D K, E D AdK, En D RdŒx�2n. utThe following corollary rephrases Theorem 17.7 in terms of sequences.

Corollary 17.8 Let s D .s˛/˛2Nd0be a real multisequence. If for each n 2 N

the truncation s.n/ WD .s˛/˛2Nd0;2n

has a representing measure supported on K, thesequence s does as well.

17.2 The Truncated Moment Problem on Projective Space

In this section, we study truncated moment problems on the real projective spacePd.R/ and apply this to the truncated K-moment problem for closed sets in Rd.

Let Pd.R/ be the d-dimensional real projective space. The points of Pd.R/ areequivalence classes of .dC 1/-tuples .t0; : : : ; td/ ¤ 0 of reals under the equivalencerelation

.t0; : : : ; td/ � .t00; : : : ; t0d/ if .t0; : : : ; td/ D �.t00; : : : ;0d / (17.13)

for some � ¤ 0. The equivalence class is denoted by Œt0 W � � � W td� and t0; : : : ; td arecalled homogeneous coordinates of the point t D Œt0 W � � � W td� 2 Pd.R/: That is,Pd.R/ D .RdC1nf0g/= �. The map

' W .t1; : : : ; td/ 7! Œ1 W t1 W � � � W td� (17.14)

is an injection of Rd into Pd.R/. We identify t 2 Rd with its image '.t/ in Pd.R/.In this manner Rd becomes a subset of Pd.R/. The complement of Rd in Pd.R/ isthe hyperplane Hd1 D fŒ0 W t1 W � � � W td� 2 Pd.R/g at infinity. Note that Hd1 can beidentified with Pd�1.R/.

We denote by HdC1;2n the homogeneous polynomials from RŒx0; x1; : : : ; xd� ofdegree 2n. Recall that RdŒx�2n are the polynomials from RŒx1; : : : ; xd� of degree atmost 2n. It is not difficult to verify that the map

� W p.x0; : : : ; xd/ 7! Op.x1; : : : ; xd/ WD p.1; x1; : : : ; xd/

is a bijection of the vector spaces HdC1;2n and RdŒx�2n with inverse given by

��1 W q.x1; : : : ; xd/ 7! Lq.x0; : : : ; xd/ WD x2n0 q

�x1x0; : : : ;

xnx0

�: (17.15)

If deg.q/ D 2n, then Lq is the homogenization of q. Clearly, the mappings ˚ and˚�1 preserve the positivity of polynomials on corresponding sets.

17.2 The Truncated Moment Problem on Projective Space 421

There is a unique topology on the space Pd.R/ for which the maps

.t0; : : : ; tj�1; tjC1; : : : ; td/ 7! Œt0 W � � � W tj�1 W 1 W tjC1 W � � � W td�; j D 1; : : : ; d;

of Rd into Pd.R/ are homeomorphisms. Then Pd.R/ is a compact topologicalHausdorff space and Rd is dense in Pd.R/. In fact, Pd.R/ is even a C1-manifold.

Each homogeneous polynomial q 2 HdC1;2n can be considered as a continuousfunction, denoted by Qq, on the projective space by

Qq.t/ WD q.t0; : : : ; td/

.t20 C � � � C t2d/n; t D Œt0 W � � � W td� 2 Pd.R/: (17.16)

(Indeed, the fraction in (17.16) is invariant under the equivalence relation (17.13),so Qq.t/ is well-defined. It is easily verified that Qq.t/ is continuous on Pd.R/.)

If we replace Rd by the projective space Pd.R/ and consider homogeneous poly-nomials, then for each closed subset K of Pd.R/ truncated K-moment functionalscan be defined almost verbatim in the same manner as for closed subsets of Rd. Wewill restate the corresponding definition in Theorem 17.9.

To formulate Theorem 17.9 we assume that N is a nonempty subset of the set

f˛ D .˛0; : : : ; ˛d/ 2 NdC10 W ˛0 C ˛1 C � � � C ˛d D 2ng:

Then A WD Lin fx˛0 W ˛ 2 Ng is a subspace of the vector space HdC1;2n. Further,suppose that K is a closed subset of the projective space Pd.R/ and define

Pos.A;K/ D f p 2 A W Qp.x/ � 0 for x 2 Kg:

Theorem 17.9 Let L be a linear functional on A. Assume that there exists an e 2 Asuch that Qe.x/ � 1 for x 2 K. Then L is a truncated K-moment functional, that is,there exists a Radon measure � on Pd.R/ with support in K such that

L. p/ DZPd.R/

Qp.t/ d�.t/ for p 2 A; (17.17)

if and only if

L. p/ � 0 for all p 2 Pos.A;K/: (17.18)

In this case, the set ML;K of such measures � is compact in the vague topology ofMC.K/ and L has a k-atomic representing measure, where k � jNj D dimA andall atoms of � are in K.

Proof The proof is almost verbatim the same as the proof of Theorem 17.3. Notethat the closed subset K of the compact space Pd.R/ is compact. The assumptionon e is needed in order to apply Proposition 1.26 with X D K. ut


It is well-known that the real projective space Pd.R/ can be identified with thequotient space Sd= � of the unit sphere

Sd D f.x0; : : : ; xd/ 2 RdC1 W x20 C x21 � � � C x2d D 1g

under the equivalence relation “�” on Sd W “x � y if and only if x D y or x D �y”.The space Sd=� is obtained from Sd by identifying antipodal points. (The map x 7!xkxk�1 gives a topological homeomorphism of RdC1=� on Sd=�.) One could eventake Sd=� with the quotient topology as the definition of the space Pd.R/.

Let SdC denote the set of points of Sd for which the first nonzero coordinate ispositive. Clearly, there is a canonical bijection of Sd=� on SdC. We can equip SdCwith the compact topology induced from Sd=� under this bijection and treat thetruncated projective moment problem on SdC. This topology on SdC is different fromthe topology induced from Sd. But, since moment functionals on finite-dimensionalspaces have finitely atomic representing measures, this does not cause any difficulty.

Now we avoid the use of the projective space Pd.R/ and look directly for integralrepresentations by measures on Sd. The next theorem restates the assertions ofTheorem 17.9 in a slightly different form.

Theorem 17.10 Let A be as above and let K be a closed subset of Sd: For a linearfunctional L ¤ 0 on A the following statements are equivalent:

(i) L. p/ � 0 for all p in Pos.A;K/ � f f 2 A W f .x/ � 0 for x 2 Kg.(ii) L is a truncated K-moment functional, that is, there exists a measure � 2

MC.Sd/ supported on K such that

L. p/ DZSdp.x/ d�.x/ for p 2 A: (17.19)

(iii) There is a k-atomic measure � 2 MC.Sd/, k � dimA, with all atoms in Kfor which (17.19) holds, that is, there are points x1; : : : ; xk 2 K and positivenumbers c1; : : : ; ck such that

L. p/ DkX

jD1cjp.xj/ for p 2 A: (17.20)

Proof Obviously, (iii)!(ii)!(i). The main implication (i)!(iii) follows fromProposition 1.26, applied to X D K and the subspace E WD AdX of C.X IR/: utRemark 17.11 Obviously, the polynomial e.x/ WD .x20C� � �Cx2d/

n satisfies Qe.t/ D 1for t 2 Pd.R/ and e.x/ D 1 for x 2 Sd. Therefore, if e is in A, it can be taken inTheorems 17.9 and 17.10 to satisfy assumption (17.1). ı

In the remaining part of this section we return to the truncated K-momentproblem for A D RdŒx�2n and study the case of a closed subset K of Rd.

17.2 The Truncated Moment Problem on Projective Space 423

We consider K as a subset of the projective space Pd.R/ by the injection 'defined by (17.14). Let K� denote the closure of K in Pd.R/. Then K� is thedisjoint union of K and the intersection K1 of K� with Hd1.

For p 2 RdŒx�2n let p2n denote its homogeneous part of degree 2n and put

fp2n.t/ WD p2n.t1; � � � ; td/.t21 C � � � C t2d/

nfor t D .0 W t1 W � � � W td/ 2 Hd1 Š Pd�1.R/:

Note that the fraction is a well-defined continuous function on Hd1:

Lemma 17.12 Let �1 be a Radon measure onHd1 supported on K1. Then

L1. p/ WDZK1

fp2n.t/ d�1.t/; p 2 RdŒx�2n; (17.21)

defines a linear functional L1 on RdŒx�2n such that L1. p/ � 0 for p 2 Pos.K/2n:

Proof Let p 2 Pos.K/2n: Since Lp.1; x/ D p.x/ � 0 for x 2 K, Lp � 0 on '.K/ Š K.Therefore, Lp � 0 on the closure K� D K [K1 of K, so that Lp.0; t/ D p2n.t/ � 0and hence fp2n.t/ � 0 for .0; t/ 2 K1: Then L1. p/ � 0 by (17.21). ut

This functional L1 satisfying condition (17.11) is the main new ingredient of thenext theorem. The following result characterizes those linear functionals on RdŒx�2nfor which the positivity condition (17.11) is fulfilled.

Theorem 17.13 Let K be a closed subset of Rd and let L be a linear functional onRdŒx�2n. Then L satisfies (17.11) if and only if there are Radon measures � on Rd

with support in K and �1 onHd1 with support in K1 such that

L. p/ DZKp.t/ d�.t/C

ZK1

fp2n.t/ d�1.t/ for p 2 RdŒx�2n: (17.22)

Proof Using Lemma 17.12 we conclude that both summands of (17.22) arenonnegative on Pos.K/2n, so L satisfies condition (17.11).

Conversely, assume that (17.11) holds. Define a linear functional LL on HdC1;2nby

LL.Lp/ D L. p/; p 2 RdŒx�2n: (17.23)

Suppose that eLp � 0 on K�: Then we have p � 0 on K by (17.15) and (17.16) andhence LL.Lp/ D L. p/ � 0 by (17.11). That is, Theorem 17.9 applies to the compactsubset X WD K� of Pd.R/ and the functional LL on E WD HdC1;2ndK�. Therefore,there exists a Radon measure Q� on Pd.R/ supported on K� such that

LL.Lp/ DZK�

eLp .t/ d Q�; p 2 RdŒx�2n: (17.24)


Define Radon measures �1 on Hd1 by �1.M/ D Q�.M/;M Hd1; and � on Rd

by

d�.t1; : : : ; td/ D .1C t21 C � � � C t2d/�nd Q�.Œ1 W t1 W � � � W td�/: (17.25)

Since supp Q� K�, �1 and � are supported on K1 and K, respectively.First let q 2 RdŒx�2n�1. Then, by (17.15) and (17.16), we have

eLq .t/ D t2n0 q� t1t0; : : : ; tdt0

�.t20 C � � � C t2d/

nD t0 q0

�t1; : : : ; td/

.t20 C � � � C t2d/n; t D Œt0 W � � � W td� 2 Pd.R/;

for some polynomial q0. Therefore, if t 2 K1, then t0 D 0, so that

eLq .t/ D 0 for t 2 K1: (17.26)

Now let p 2 RdŒx�2n. Then q WD p � p2n 2 RdŒx�2n�1: Again by (17.15)and (17.16),

eLp .t/ D p�t1; : : : ; td/

.1C t21 C � � � C t2d/n

for t D Œ1 W t1 W � � � W td� 2 K: (17.27)

Hence, using the formulas (17.23)– (17.27) we derive

L. p/ D LL.Lp/ DZK�

eLp .t/ d Q�.t/ DZKeLp .t/ d Q�.t/C

ZK1

fp2n.t/ d Q�.t/CZK1

eLq .t/ d Q�.t/

DZK

p.t1; : : : ; td/

.1C t21 C � � � C t2d/nd Q�.Œ1 W t1 W � � � W td�/C

ZK1

fp2n.t/ d�1.t/

DZKp.t/ d�.t/C

ZK1

fp2n.t/ d�1.t/;

which proves (17.22). utRemark 17.14

1. If the set K in Theorem 17.13 is compact, then K1 is empty, so the secondsummand in (17.22) does not occur and we obtain Theorem 17.6.

2. Theorem 17.13 explains why for noncompact sets K the positivity condi-tion (17.11) is not sufficient for L to be a truncated K-moment functional. Thereason is that there may be a functional L1 given by some measure at “infinity”.

3. Most of the results and notions developed in this chapter and the next have theircounterparts for the truncated moment problem on Pd.R/. We leave it to thereader to state the corresponding projective versions.

4. The truncated moment problem on a closed subset K of Pd.R/ has severaladvantages. First, since Pd.R/ and hence K is compact, truncated K-moment

17.3 Hankel Matrices 425

functionals are characterized by the positivity condition (17.18). As noted inRemark 2, for the truncated moment problem on Rd condition (17.18) is notsufficient. Secondly, homogeneous polynomials are more convenient to deal withand additional technical tools such as the apolar scalar product (see Sect. 19.1)are available. ıFinally, we use the truncated moment problem on Pd.R/ to derive the following

result on the truncated moment problem for closed subsets of Rd.

Theorem 17.15 Let K be a closed subset of Rd and L0 a linear functional onRdŒx�2n�2; n 2 N. Then L0 is a truncated K-moment functional on RdŒx�2n�2 ifand only if L0 admits an extension to a linear functional L on RdŒx�2n such that

L. p/ � 0 for p 2 Pos.K/2n: (17.28)

Proof Let L0 be a truncated K-moment functional. Then, by Theorem 17.2, L0 hasa finitely atomic representing measure �. Clearly, RdŒx�2n L1.K; �/ and thefunctional L defined by L. p/ D R f d�, p 2 RdŒx�2n, satisfies (17.28).

Conversely, assume that (17.28) holds. Then, by Theorem 17.13, L is of theform (17.22). If p 2 RdŒx�2n�2, then p2n D 0. Hence the second summand in (17.22)vanishes and we obtain L0. p/ D

RK p.t/ d�.t/ for p 2 RdŒx�2n�2. ut

17.3 Hankel Matrices

Recall that N is a nonempty subset of Nd0 and A D Lin fx˛ W ˛ 2 Ng: Throughout

this section, we suppose that L is a linear functional on the linear subspace

A2 WD Lin f p q W p; q 2 Ag D Lin fxˇ W ˇ 2 NC Ng

of RdŒx�. Note that in contrast to Sects. 17.1–17.2 we now consider functionals onA2 rather than A:

The following definition contains the basic notions studied in this section.

Definition 17.16 The Hankel matrix of L is the symmetric matrix

H.L/ D .h˛;ˇ/˛;ˇ2N; where h˛;ˇ WD L.x˛Cˇ/; ˛; ˇ 2 N:

The (cardinal) number rankL WD rankH.L/ is called the rank of L. The kernel NL

and the real algebraic set VL are defined by

NL D f f 2 A W L. fg/ D 0 for all g 2 Ag; (17.29)

VL D ft 2 Rd W f .t/ D 0 for all f 2 NLg: (17.30)


Hankel matrices are fundamental tools for the study of truncated momentproblems. Many problems and properties of the functional L are easily translatedinto those for the matrix H.L/, see e.g. Proposition 17.17 below.

Further, we introduce a symmetric bilinear form h�; �i0 on A � A by

h f ; gi0 WD L. fg/; f ; g 2 A:

Equation (17.29) means that NL is just the kernel of the bilinear form h �; �i0: Thisis one reason for the importance of the vector space NL. We define a symmetricbilinear form h �; �i on the quotient space DL WD A=NL by

h f CNL; gCNLi WD h f ; gi0 D L. fg/; f ; g 2 A: (17.31)

Clearly, this bilinear form h �; �i is non-degenerate, that is, its kernel is trivial. Let uswrite Qf for the equivalence class f CNL. Then, by (17.31) we have

L. fg/ D hQf ; Qgi; f ; g 2 A: (17.32)

Since each element of A2 is a sum of products fg with f ; g 2 A, all numbers s˛ ,˛ 2 NC N; and hence the functional L can be recovered from the space .DL; h�; �i/by using Eq. (17.32).

Let us fix an ordering of the index set N; for instance, we can take thelexicographic ordering. For f D P

˛2N f˛x˛ 2 A we let Ef D . f˛/T denote the

coefficient vector of f written as a column according to the ordering of N. SinceEf has only finitely many nonzero terms f˛ , products such as Ef TH.L/ and H.L/Ef arewell-defined. If the set N is finite, then Ef 2 RjNj:

A number of basic facts on Hankel matrices are collected in the next proposition.In dimension one some assertions have been already noted in Lemma 9.20.

Proposition 17.17

(i) For f 2 A and g 2 A we have

L. fg/ D Ef TH.L/Eg: (17.33)

(ii) A polynomial f 2 A belongs to f 2 NL if and only if Ef 2 kerH.L/: Inparticular, dimNL D dim kerH.L/.

(iii) rankL � rankH.L/ D dim .A=NL/ � dim DL.(iv) L is a positive functional (that is, L. f 2/ � 0 for f 2 A by Definition 2.2) if and

only if the Hankel matrix H.L/ is positive semidefinite.(v) If the functional L is positive, thenNL D f f 2 A W L. f 2/ D 0g.

(vi) If L is a truncated moment functional, then supp� VL for each representingmeasure � of L:

17.3 Hankel Matrices 427

Proof

(i) Let f D P˛2N f˛x

˛ 2 A and g D P˛2N g˛x

˛ 2 A. Clearly, then we havefg DP˛;ˇ2N f˛gˇx

˛Cˇ and therefore

L. fg/ DX˛;ˇ2N

f˛gˇL.x˛Cˇ/ D

X˛;ˇ2N

h˛;ˇf˛gˇ D . f˛/ TH.L/.gˇ/ D Ef TH.L/Eg:

(ii) follows at once by combining the definition (17.29) of NL and (17.33).(iii) Let us take m 2 N column vectors

h.1/ D .h˛.1/;ˇ/ˇ2N; : : : ; h.m/ D .h˛.m/;ˇ/ˇ2N; where ˛.1/; : : : ; ˛.m/ 2 N;

of the Hankel matrix H.L/ and consider a linear combination of these vectorswith real coefficients c1; : : : ; cm. Put p.x/ DPi cix

˛.i/ . Then p 2 A and

mXiD1

cih˛.i/;ˇ D L

� mXiD1

cix˛.i/xˇ

�D L. p.x/xˇ/; ˇ 2 N:

Hence this linear combination of h.1/; : : : ; h.m/ is zero if and only if p 2NL: Thus, fh.1/; : : : ; h.m/g is a maximal set of linearly independent columnvectors of H.L/ if and only if fx˛.1/ ; : : : ; x˛.m/g is a maximal set of linearlyindependent monomials in the quotient vector space DL D A=NL. This impliesrankH.L/ D dim DL.

If the set N is finite, this equality can be obtained by a shorter reasoning:Using that dim kerH.L/ D dimNL by (ii) and the rank-nullity characterizationwe get

rankH.L/ D dimRjNj � dim kerH.L/

D dim A� dimNL D dim.A=NL/ D dim.DL/:

(iv) is obtained from (17.33) by setting f D g.(v) If f 2 NL, then L. f 2/ D 0 by setting f D g in (17.33). We prove the converse.

Since L is a positive functional, the Cauchy–Schwarz inequality (2.7) holds:

L. fg/2 � L. f 2/L.g2/ for f ; g 2 A: (17.34)

Therefore, if L. f 2/ D 0, then L. fg/ D 0 for all g 2 A, so that f 2 NL.(vi) follows at once from Proposition 1.23. ut

Suppose that the functional L is positive on A2, that is, L. f 2/ � 0 for f 2 A:Then, by (17.32), the non-degenerate symmetric bilinear form h�; �i onDL is positivedefinite and hence a scalar product. Thus, .DL; h�; �i/ is a real unitary space. Further,


by Proposition 17.17(iii), DL is finite-dimensional if and only if the Hankel matrixH.L/ has finite rank. In this case the unitary space .DL; h�; �i/ is obviously complete,that is, .DL; h�; �i/ is a real finite-dimensional Hilbert space. In particular, DL hasfinite dimension if the set N is finite.

The next proposition shows that NL obeys some ideal-like properties.

Proposition 17.18 Let p 2 NL and q 2 A. Suppose that pq 2 A.

(i) If L is a positive functional and pq2 2 A, then pq 2 NL.(ii) If L is a truncated K-moment functional, then pq 2 NL.

Proof

(i) Since the functional L is positive, the Cauchy–Schwarz inequality (17.34)holds. Using that pq2 2 A and L. p2/ D 0 we obtain

L.. pq/2/2 D L. p pq2/2 � L. p2/L.. pq2/2/ D 0;

so that pq 2 NL by Proposition 17.17(iv).(ii) Let � 2ML;K. Recall that supp� VL by Proposition 17.17(vi). Therefore,

since p 2 NL and hence p.x/ D 0 on VL, we get

L.. pq/2/ DZK. pq/2.x/ d�.x/ D

ZVL\K

p.x/2q.x/2 d�.x/ D 0:

Thus, p q 2 NL again by Proposition 17.17(iv). utWe restate the preceding results in the case of our standard example.

Corollary 17.19 Let N be the set Nd0;2n (see (17.9)) and let L be a linear functional

on A D RdŒx�2n. Suppose that p 2 NL and q 2 RdŒx�n.

(i) If L is a positive functional and p q 2 RdŒx�n�1, then p q 2 NL.(ii) If L is a truncated K-moment functional and p q 2 RdŒx�n, then p q 2 NL.

Proof

(i) Proposition 17.18(i) yields the assertion in the case q D xj; j D 1; : : : ; d. SinceNL is a vector space, repeated applications give the general case.

(ii) follows from Proposition 17.18(ii). utRemark 17.20 Example 9.29, Case 1, shows that the assertion p q 2 NL inCorollary 17.19(i) is no longer valid if p q 2 RdŒx�n. This provides an importantdifference between positive functionals and moment functionals! ı

17.4 Hankel Matrices of Functionals with Finitely Atomic Measures 429

17.4 Hankel Matrices of Functionals with FinitelyAtomic Measures

Throughout this section, � is a finitely atomic signed measure

� DkX

jD1mjıxj ; where mj 2 R; xj 2 Rd for j D 1; : : : ; k; (17.35)

and L denotes the corresponding functional on A2 defined by

L. f / DZ

f .x/ d� DkX

jD1mjf .xj/; f 2 A2: (17.36)

Ifmj � 0 for all j, then� is a Radon measure and L is a truncated moment functional.Clearly, � is k-atomic if all mj > 0 and the points xj are pairwise distinct.

For x 2 Rd, we denote the column vector .x˛/˛2N by sN.x/ or simply by s.x/ ifno confusion can arise. Note that sN.x/ is the moment vector of the delta measure ıxfor A, not for A2!

Proposition 17.21 For the Hankel matrix H.L/ of the functional L we have

H.L/ DkX

jD1mjsN.xj/ sN.xj/

T ; (17.37)

rankH.L/ � k D jsupp�j � jVLj: (17.38)

Suppose that mj ¤ 0 for j D 1; : : : ; k. Then the following are equivalent:(i) rankH.L/ D k.

(ii) The point evaluations lx1 ; : : : ; lxk on A (!) are linearly independent.(iii) The vectors sN.x1/; : : : ; sN.xk/ are linearly independent.

Proof Clearly, for x 2 Rd the .˛; ˇ/-entry of the matrix s.x/s.x/T is just x˛Cˇ:This means that s.x/s.x/T is the Hankel matrix of the point evaluation lx on A2.Since L DPj mjlxj , this yields the formula (17.37) for H.L/.

By (17.37), the matrix H.L/ is a sum of k matrices mjs.xj/s.xj/T . These matricesare of rank one (if mj ¤ 0) or rank zero (if mj D 0). Hence rankH.L/ � k. Sincesupp� VL, it is obvious that jsupp�j � jVLj:

Now we assume that mj ¤ 0 for all j and prove the equivalence of (i)–(iii).

(i)$(ii) Let f D P˛2N f˛x

˛ 2 A: Recall that Ef is the column vector . f˛/T : Leth.˛/ be the ˛-th column of H.L/ and e˛ D .ı˛;ˇ/ˇ2N the ˛-th basis vector. Then

H.L/Ef D X˛2N

f˛H.L/e˛ D X˛2N

f˛h.˛/ D X

˛2N

f˛

kXjD1

mjx˛j s.xj/ D

kXjD1

mjlxj. f /s.xj/:

(17.39)


Since mj ¤ 0 and imH.L/ is contained in the span of vectors s.x1/; : : : ; s.xk/,it follows from (17.39) that rankH.L/ D dim imH.L/ is k if and only if therestrictions of point evaluations lx1 ; : : : ; lxk to A are linearly independent.

(ii)$(iii) Clearly, s.xj/ is the column .lxj.x˛//˛2N. Thus, if c1; : : : ; ck 2 R, thenP

j cjs.xj/ is the column vector�.P

j cjlxj�.x˛//˛2N: Therefore,

Pj cjs.xj/ D 0 if and

only ifP

j cjlxj D 0 on A. Hence (ii) and (iii) are equivalent. utCorollary 17.22 For each truncated K-moment functional L on A2 we have

rankH.L/ � jVL \Kj: (17.40)

Proof By Theorem 17.2 and Proposition 17.17(vi), there is a k-atomic measure� 2 ML;K and supp� VL \ K: Since rankH.L/ � jsupp�j by (17.38), thisyields (17.40). utExample 17.23 Let d D 1;N D f0; 1g, so that A D fa C bx W a; b 2 Rg. Thefunctionals l�1; l0; l1 are linearly independent on A2, but they are linearly dependenton A, since 2l0 D l�1 C l1 on A. For L D l�1 C l0 C l1 we have rankH.L/ D 2. ı

Note that (17.40) is a necessary condition for truncated K-moment functionals.There are several necessary conditions for a linear functional L on A2 to be a

truncated moment functional. These are� the positivity condition:

L. f 2/ � 0 for f 2 A; (17.41)

� the rank-variety condition:

rankH.L/ � jVLj; (17.42)

� the consistency condition:

p 2 NL; q 2 A and pq 2 A imply pq 2 NL: (17.43)

The positivity condition is obvious. Conditions (17.42) and (17.43) followfrom (17.40) and Proposition 17.18(ii), applied with K D Rd.

Another necessary condition is given in Proposition 18.15(ii) below.

Remark 17.24 Let h.˛/ D .h˛;ˇ/ˇ2N denote the ˛-th column vector and CH.L/ thelinear span of column vectors h.˛/, ˛ 2 N, of the Hankel matrix H.L/. We define asurjective linear mapping ' W A! CH.L/ by

f DX˛2N

f˛x˛ 2 A 7! '. f / D

X˛2N

f˛h.˛/:

17.4 Hankel Matrices of Functionals with Finitely Atomic Measures 431

These “functions” '. f / of column vectors are another tool. By (17.39) we have

H.L/Ef DX˛2N

f˛h.˛/ D '. f / for f D

X˛2N

f˛x˛ 2 A:

Hence rankH.L/ D dim imH.L/ D dimCH.L/: By Proposition 17.17(ii), NL is thekernel of '. Hence the consistency condition (17.43) can be reformulated as

p; q; p q 2 A and '. p/ D 0 H) '. p q/ D 0: ı

Example 17.28 below shows that it may happen that jVLj > rankH.L/. Now weturn to functionals for which equality holds in the rank-variety condition (17.42).

Definition 17.25 A linear functional L on A2 is called minimal if rankH.L/ is finiteand rankH.L/ D jVLj:

Our next aim is to derive two simple but very useful formulas (17.45)and (17.47).

Let � and L be as above and let F D f f1; : : : ; fng be a finite subset of A: Define

MF D

[email protected]/ f1.x2/ : : : f1.xk/f2.x1/ f2.x2/ : : : f2.xk/: : : : : : : : : : : :

fn.x1/ fn.x2/ : : : fn.xk/

1CCA �

0BB@lx1 . f1/ lx2 . f1/ : : : lxk. f1/lx1 . f2/ lx2 . f2/ : : : lxk. f2/: : : : : : : : : : : :

lx1 . fn/ lx2 . fn/ : : : lxk. fn/

1CCA :

Then, using (17.36) we compute

MF .m1; : : : ;mk/T D .L. f1/; : : : ;L. fn//T : (17.44)

For minimal functionals L we will use (17.44) to derive the masses mj fromthe points xi and the functional L. If L is minimal, then rankH.L/ D jVLj andhence (17.38) and Proposition 17.33(iii) imply that k D rankH.L/ D dim.A=NL/.

Lemma 17.26 Let � and L be defined by (17.35) and (17.36), respectively,and suppose that L is minimal. Let f f1; : : : ; fkg be a subset of A such that therepresentatives Qfj D fj C NL form a basis of the quotient space A=NL. Then thek � k-matrix MF is invertible and

.m1; : : : ;mk/T D .MF /�1.L. f1/; : : : ;L. fk//T : (17.45)

Proof Assume to the contrary that the matrix MF is not invertible. Then the columnrank of MF is less than k, so there exist reals ˛1; : : : ; ˛k, not all zero, such that˛1f1.xj/C � � � C ˛kfk.xj/ D 0 for all j. From the definition (17.36) of L we computethat f WD ˛1f1 C � � � C ˛kfk 2 A satisfies L. f 2/ D 0, so that f 2 NL: Thiscontradicts the linear independence of the set f Qf1; : : : ; Qfkg in A=NL. Thus, MF isinvertible and (17.45) follows from (17.44). ut


Now we introduce the matrix

HF .L/ D

0BB@L. f 21 / L. f1f2/ : : : L. f1fn/L. f1f2/ L. f 22 / : : : L. f2fn/: : : : : : : : : : : :

L. f1fn/ L. f2fn/ : : : L. f 2n /

1CCA : (17.46)

If N is finite and F is our standard basis fx˛ W ˛ 2 Ng, then HF .L/ is just the “usual”Hankel matrix H.L/ from Definition 17.16. Let D.m/ denote the diagonal matrixwith diagonal entries m1; : : : ;mk. Then a simple computation yields

0BB@L. f 21 / L. f1f2/ : : : L. f1fn/L. f1f2/ L. f 22 / : : : L. f2fn/: : : : : : : : : : : :

L. f1fn/ L. f2fn/ : : : L. f 2n /

1CCA D

[email protected]/ f1.x2/ : : : f1.xk/f2.x1/ f2.x2/ : : : f2.xk/: : : : : : : : : : : :

fn.x1/ fn.x2/ : : : fn.xk/

1CCA

0BB@m1 0 : : : 0

0 m2 : : : 0

: : : : : : : : : : : :

0 0 : : : mk

1CCA

[email protected]/ f2.x1/ : : : fn.x1/f1.x2/ f2.x2/ : : : fn.x2/: : : : : : : : : : : :

f1.xk/ f2.xk/ : : : fn.xk/

1CCA

which means that

HF .L/ D MFD.m/.MF /T : (17.47)

Proposition 17.27 If N is finite and L is a minimal truncated moment functional onA2, then L is determinate and its unique representing measure is rankH.L/-atomic.

Proof Set k WD rankH.L/. Since L is minimal, k D jVLj. We write VL Dfx1; : : : ; xkg. Let � be an arbitrary representing measure of L. Since supp � VL

by Proposition 17.17(vi), � is of the form � DPkjD1mjıxj with mj � 0 for all j.

Now we apply (17.47) to the standard Hankel matrix H.L/. If mj D 0 for one j,then rankD.m/ < k and hence rankH.L/ < k by (17.47), which is a contradiction.Thus, mj > 0 for all j, so � is k-atomic.

We choose a set f f1; : : : ; fkg as in Lemma 17.26. Then the numbers mj are givenby (17.45), so the measure � is uniquely determined and L is determinate. utExample 17.28 (An example for which jVLj > rankH.L/) Suppose that d D 2;

n � 3; and A D RŒx1; x2�n: Set

p.x1; x2/ D .x1 � ˛1/ � � � .x1 � ˛n/; q.x1; x2/ D .x2 � ˇ1/ � � � .x2 � ˇn/;

where ˛i; ˇj are real numbers such that ˛1 < � � � < ˛n and ˇ1 < � � � < ˇn: ThenZ. p/\ Z.q/ D f.˛i; ˇj/ W i; j D 1; : : : ; ng.

17.5 The Full Moment Problem with Finite Rank Hankel Matrix 433

We define an n2-atomic measure � such that all atoms are in Z. p/ \ Z.q/ anda truncated moment functional L on A2 by L. f / D R

f d�; f 2 A2. Then we haveL. p2/ D L.q2/ D 0, so that p; q 2 NL. Clearly, p and q are linearly independent inNL. Hence rankH.L/ D dim .A=NL/ �

�nC22

��2. From supp� VL and p; q 2 NL

we obtain n2 D jsupp�j � jVLj � jZ. p/\ Z.q/j D n2. Thus, VL D Z. p/\Z.q/and

jVLj � rankH.L/ � n2 � nC 22

!C 2 D

n � 12

!: (17.48)

We show that L is determinate. Let � be a representing measure of L. Sincesupp � VL D Z. p/ \ Z.q/, � can be written as � D Pn

i;jD1 mijı.˛i;ˇj/: Let us fixi; j 2 f1; : : : ; ng: Put fij WD .x1 � ˛i/�1.x2 � ˇj/�1p q. Then L. fij/ D

Rfij d� D mij.

Hence the measure � is uniquely determined by L, that is, L is determinate.From the assumption n � 3 and (17.48) it follows that jVLj > rankH.L/ and the

difference jVLj � rankH.L/ is large as n becomes large. Since L is determinate, Lhas no .rankL/-atomic representing measure, so L is not minimal. ı

17.5 The Full Moment Problem with Finite RankHankel Matrix

Throughout this section, we assume that N D Nd0. Then A D A2 D RdŒx� and we are

concerned with the full moment problem.Let L be a positive functional on the real �-algebra RdŒx� with identity involution

(that is, f � D f for f 2 RdŒx�). We consider the GNS construction associatedwith L, see Sect. 12.5. By Proposition 17.17(iv), the vector space NL definedin (17.29) coincides with the left ideal NL from Lemma 12.38. Therefore, by(17.32), .DL; h�; �i/ is the domain of the GNS representation �L of RdŒx�. Recallthat, by the definition of the GNS representation,�L. p/ acts on DL by multiplicationwith p.

Theorem 17.29 Suppose that L is a positive linear functional on RdŒx� such thatrankH.L/ D r, where r 2 N. Then L is a moment functional with uniquerepresenting measure �. This measure � has r atoms and supp� D VL.

Proof By Proposition 17.17, dim DL D rankH.L/ D r. As noted before thetheorem, DL is the representation space of the GNS representation �L. Therefore,Xk WD �L.xk/, k D 1; : : : ; d; are commuting self-adjoint operators acting on the r-dimensional unitary space DL. From linear algebra it is known that these operatorshave a common set of eigenvectors, say e1; : : : ; er, which form an orthonormal basisof DL. Let Xkej D tkjej and put tj D .t1j; : : : ; tdj/T . Since all eigenvalues of Xk arereal, tj 2 Rd. We develop the vector 1 2 DL with respect to this orthonormal basis


and obtain 1 DPrjD1mjej. Set � WDPr

jD1m2j ıtj . For p 2 RdŒx� we derive

L. p/ D h�L. p/1; 1i D hp.�L.x1/; : : : ; �L.xd//1; 1i D hp.X1; : : : ;Xd/1; 1i

DrX

i;jD1hp.t1j; : : : ; tdj/mjej;mieii D

rXjD1

p.tj/m2j D

Zp.t/ d�.t/: (17.49)

Here the first equality holds by formula (12.33), while the second equality followsfrom the fact that �L is an algebra homomorphism. (This is just the finite-dimensional version of the proof of Theorem 12.40.) By (17.49),� is a representingmeasure of L.

By the properties of the GNS construction, �L.RdŒx�/1 D DL. Hence mj ¤ 0 forall j. Since rankH.L/ D r, the point evaluations lt1 : : : ; ltr on A D RdŒx� are linearlyindependent by Proposition 17.21. Therefore, ti ¤ tj for i ¤ j. By the preceding wehave shown that � DPr

jD1m2j ıtj is r-atomic. Because � has finite, hence compact,support, Proposition 12.17 implies that L has only one representing measure. (Thisfact also follows from the uniqueness of the spectral decomposition of X1; : : : ;Xd.)

By Proposition 17.17(vi), supp� VL. We prove that VL supp�. Let t0 2 VL.Assume to the contrary that t0 … supp� D ft1; : : : ; trg: By Lagrange interpolationthere exists a p 2 RdŒx� such that p.t0/ D 1 and p.tj/ D 0 for j D 1; : : : ; r. Then

L. p2/ DZ

p2 d� DrX

jD1mjp.tj/

2 D 0;

so that p 2 NL. Since p.t0/ D 1, t0 … VL, which is a contradiction. Thus we haveshown that VL D supp�. ut

17.6 Flat Extensions and the Flat Extension Theorem

In this section we derive another result, called the flat extension theorem, whichprovides a sufficient condition for the solvability of the truncated moment problem.To formulate this theorem we begin with some preliminaries.

We suppose in this section that N is a finite subset of Nd0, A D fx˛ W ˛ 2 Ng

and A D Linfx˛ W ˛ 2 Ng. Now let N0 be another subset of Nd0. We denote by

B D fx˛ W ˛ 2 N0g the corresponding monomials and by B the linear span ofelements of B.

Definition 17.30 We say the set B is connected to 1 if 1 2 B and each p 2 B,p ¤ 1, can be written as p D xi1 : : : xik with xi1 ; xi1xi2 ; : : : ; xi1 � � � xik 2 B:

Example 17.31 Clearly, f1; x1; x2; x1x2g and f1; x2; x2x3; x1x2x3g are connected to 1,but f1; x1; x2; x1x2x3g is not. ı

17.6 Flat Extensions and the Flat Extension Theorem 435

From now on we assume that B is a subset of A. Then B2 is a subspace of A2:

Definition 17.32 A linear functional L on A2 is called flat with respect to B if

rankH.L/ D rankH.L0/;

where L0 denotes the restriction to B2 of L.

To motivate this definition we write the Hankel matrix H.L/ as a block matrix

H.L/ D�H.L0/ X12X21 X22

�:

The matrix H.L/ is called a flat extension (see Definition A.22) of the matrix H.L0/if rankH.L/ D rankH.L0/. Hence the functional L is flat with respect to B if andonly if the Hankel matrix H.L/ is a flat extension of the Hankel submatrix H.L0/.

The next proposition contains a reformulation of the flatness condition that can beused in noncommutative settings as well. Another proof based on Hankel matricesis sketched in Exercise 17.6. Recall that NL is defined by (17.29).

Proposition 17.33 A linear functional L on A2 is flat with respect to B if and onlyif A D BCNL. In this case, NL0 D NL \ B:

Proof Clearly, NL \ B NL0 and there are canonical maps �1 and �2:

B=NL0�1 B=.NL \ B/

�2! A=NL:

Here the dimensions are increasing from left to right, so by Proposition 17.17(iii),

rankH.L0/ D dim .B=NL0 / � dim .B=.NL \ B/

� dim .A=NL/ D rankH.L/:

Therefore, L is flat with respect to B if and only if we have equality throughout, thatis, �1 and �2 are bijections, or equivalently, A D BCNL and NL0 D NL \ B:

To complete the proof we show that A D B C NL implies that NL0 D NL \ B.Obviously, NL \ B NL0 . To prove the converse inclusion, let f 2 NL0 . We writeg 2 A as g D g1 C g2 with g1 2 B and g2 2 NL. Then fg1 2 B2. Since f 2 NL0 andg2 2 NL, we get L. fg/ D L. fg1/C L. fg2/ D L0. fg1/ D 0, so that f 2 NL \ B: utProposition 17.34 Let L be a linear functional on A2 which is flat with respect toB. If L. p2/ � 0 for all p 2 B2, then L.q2/ � 0 for all q 2 A2.

Proof Let L0 denote the restriction to B2 of L. Since L0 is positive on B2 byassumption, H.L0/ � 0. Because L is flat with respect to B, H.L/ is a flat extensionof H.L0/. Then H.L/ � 0 by Theorem A.24(iii). Hence L is positive on A2. ut

The next result is the flat extension theorem.


Theorem 17.35 Suppose that B is a finite set of monomials of RdŒx� such that B isconnected to 1 and

A D B [ x1�B [ � � � [ xd�B: (17.50)

Let A and B be the linear spans of A and B, respectively. Suppose that L is a linearfunctional on A2 which is flat with respect to B.

Then L has a unique extension to a linear functional QL of RdŒx� such that QL isflat with respect to A. If L. p2/ � 0 for all p 2 B, then QL. f 2/ � 0 for all f 2 RdŒx�.Further, if I.NL/ denotes the ideal of RdŒx� generated byNL, then

NQL D I.NL/ and VQL D VL: (17.51)

The functional QL on RdŒx� is also flat with respect to B, because the functional Lon A2 is flat with respect to B.

The proof of Theorem 17.35 is lengthy and will be given in the next section. Thefollowing theorem of Curto and Fialkow contains the main applications concerningthe truncated moment problem.

Theorem 17.36 Let A;B;A;B satisfy the assumptions of Theorem 17.35 and let Lbe a linear functional on A2 which is flat with respect to B. Suppose that L. p2/ � 0for p 2 B. Then L is a minimial truncated moment functional.

In particular, L is determinate and its unique representing measure� is r-atomic,where r WD rankH.L/ � jBj, and satisfies supp� D VL.

Proof Let QL be the positive linear functional on RdŒx� from Theorem 17.35. Sinceboth L and QL are flat with respect to B, we have

r D rankH.L/ D rankH. QL/ D rankH.L0/ � jBj:Therefore, it follows from Theorem 17.29 that QL, hence L, has an r-atomicrepresenting measure � and supp� D VQL. Since VL D VQL by (17.51), we haver D rankH.L/ D jVLj, so L is minimal and hence determinate by Proposition 17.27.

utThe most important application of Theorem 17.36 concerns the following case:

N0 D Nd0;n�1; N D Nd

0;n; B D fx˛ W j˛j � n � 1g; A D fx˛ W j˛j � ng:Then it is obvious that B is connected to 1, assumption (17.50) holds,

B2 D RdŒx�2n�2; and A2 D RdŒx�2n:

Let L be a linear functional on RdŒx�2n. Recall from Definition 17.16 that the Hankelmatrix of L is the matrix Hn.L/ WD H.L/ with entries

h˛;ˇ WD L.x˛Cˇ/; where ˛; ˇ 2 Nd0; j˛j; jˇj � n:

17.6 Flat Extensions and the Flat Extension Theorem 437

Clearly, the Hankel matrix of the restriction L0 of L to RdŒx�2n�2 is the matrixHn�1.L/ WD H.L0/ with entries h˛;ˇ, where ˛; ˇ 2 Nd

0; j˛j; jˇj � n � 1: Further,jBj D dimRdŒx�n�1 D

�n�1Cdd

�. Then Theorem 17.36 has the following special

case.

Theorem 17.37 Suppose that L is a linear functional on RdŒx�2n; n 2 N, such that

L. p2/ � 0 for p 2 RdŒx�n�1 and r WD rank Hn.L/ D rank Hn�1.L/: (17.52)

Then L is a determinate truncated moment functional and its unique representingmeasure is r-atomic with r � �n�1Cd

d

�.

The next theorem deals with the truncated K-moment problem, where

K WD K.f/ D fx 2 Rd W f0.x/ � 0; : : : ; fk.x/ � 0g (17.53)

is the semi-algebraic set associated to a finite subset f D f f0; : : : ; fkg of RdŒx� withf0 D 1: Let us abbreviate m WD maxf1; deg. fj/ W j D 1; : : : ; kg:Theorem 17.38 Let L be a linear functional on RdŒx�2n, n 2 N, and r WDrankHn.L/. Retaining the preceding notation, the following are equivalent:

(i) L is a truncated K-moment functional which has an r-atomic representingmeasure � with all atoms in K.

(ii) L extends to a linear functional QL on RdŒx�2.nCm/ such that rankHnCm. QL/ DrankHn.L/ and QL. fjp2/ � 0 for all p 2 RdŒx�n and j D 0; : : : ; k:

Proof(i)!(ii) Let QL be the truncated K-moment functional on RdŒx�2.nCm/ given by a

measure � as in (i). Then, by (17.38),

r D jsupp�j � rankHnCm. QL/ � rankHn.L/ D r;

so that rankHnCm. QL/ D rankHn.L/. The positivity condition is obvious.(ii)!(i) Since rankHn.L/ D rankHnCm. QL/, QL is flat with respect to RdŒx�n.

Therefore, since L. p2/ � 0 for p 2 RdŒx�n by assumption (with f0 D 1),Proposition 17.34 implies that QL.q2/ � 0 for q 2 RdŒx�nCm. Further, QL is flat withrespect to RdŒx�nCm�1: (This is where m � 1 is used!) Hence, by Theorem 17.37,QL, hence L, is a truncated moment functional with r-atomic representing measure� DPr

iD1miıxi , where all mi > 0. It remains to prove that all atoms xj; j D 1; : : : ; r;are in K.

Since jsupp�j D r D rankHn.L/, it follows from Proposition 17.21 (i)$(ii)that the functionals lx1 ; : : : ; lxr are linearly independent on RdŒx�n. Hence there arepolynomials p1; : : : ; pr 2 RdŒx�n such that lxi. pj/ � pj.xi/ D ıij for all i; j. Let


j; k 2 f0; : : : ; rg. Then we have fkp2j 2 RdŒx�2.nCm/ and

QL. fkp2j / DZ

fkp2j d� D

rXiD1

mifk.xi/pj.xi/2 D mjfk.xj/ � 0:

Since mj > 0, we conclude that fk.xj/ � 0 for all k. Hence xj 2 K by (17.53). utRemark 17.39

1. The proof of the implication (ii)!(i) shows that � is also a representing measurefor the functional QL on RdŒx�2.nCm/.

2. As we have discussed in Remark 16.10, Theorem 17.38 has an interestingapplication to polynomial optimization over the set K. ı

17.7 Proof of Theorem 17.36

The following “truncated ideal-like” property of NL will be used twice below.

Lemma 17.40 If f 2 NL and xi f 2 A, then xi f 2 NL:

Proof Since xi f 2 A, we have xi f D p C q with p 2 B and q 2 NL by theequality A D B C NL from Proposition 17.33. Therefore, by f 2 NL and by theassumption (17.50), for all g 2 B A we have xig 2 xiB A and

0 D L. f .xig// D L. pg/C L.qg/ D L. pg/ D L0. pg/;

so that p 2 NL0 . Hence p 2 NL and xi f D pC q 2 NL. utBy Proposition 17.33 the flatness implies that A D BCNL. We choose a subspace

D of B such that A is the direct sum of vector spaces D and NL.Without loss of generality we assume that 1 2 D. (If such a choice is impossible,

then 1 2 NL. But then a repeated application of Lemma 17.40 yields A 2 NL, sothat L � 0 and the assertion holds trivially.)

Let � denote the projection of A onto D B with respect to the direct sum

A D D ˚NL: (17.54)

Further, let Xi W D ! D; i D 1; : : : ; d, denote the operator Xi W D ! D defined byXi. f / WD �.xi f /, f 2 D. The crucial step of the proof is the following

Lemma 17.41 If f 2 D, then XiXjf D XjXi f for i; j D 1; : : : ; d and f 2 D.

Proof Let f 2 D. By the definition of the operators Xj and Xi we obtain

XiXj. f / D Xi.�.xjf // D �.xi�.xjf // D xixjf � xi.I��/.xjf / � .I��/.xi.�.xjf ///:

17.7 Proof of Theorem 17.36 439

Therefore, .XiXj � XjXi/f D g0 C g1 C g2, where

g0 WD .I � �/.xj.�.xi f ///� .I � �/.xi.�.xjf ///;g1 WD xj.I � �/.xi f /; g2 WD �xi.I � �/.xjf /:

Clearly, g0 2 NL by the definition of the projection � . Since .I � �/.xi f / 2 NL

and g1 D xj.I � �/.xjf / 2 xjB A, Lemma 17.40 implies that g1 2 NL. The samereasoning shows that g2 2 NL. Thus g0 C g1 C g2 2 NL.

On the other hand, the sum g0 C g1 C g2 D .XiXj � XjXi/f belongs to D. SinceD \NL D f0g by (17.54), we conclude that XiXjf D XjXi f . ut

Let p 2 RdŒx�. Since the operators Xi and Xj on D commute by Lemma 17.41,it follows that p.X/ WD p.X1; : : : ;Xd/ is a well-defined linear operator mapping Dinto itself. Recall that 1 2 D. Set '. p/ WD P.X/1 2 D. Clearly, the map p 7! p.X/is an algebra homorphism of RdŒx� into the linear operators of D. Hence we have

'. pq/ D p.X/'.q/; p; q 2 RdŒx�: (17.55)

In particular, this implies that ker' is an ideal of RdŒx�.

Lemma 17.42

(i) '. p/ D �. p/ for p 2 A.(ii) L. pq/ D L.'. pq// for p; q 2 A:

Proof

(i) The proof is given by induction on the degree of p. Obviously, '.1/ D �.1/ D1. Suppose that the assertion holds for all p of degree k. Let q 2 A and deg.q/ Dk C 1. It suffices to prove the assertion for monomials q 2 A. Since A D BC,we have q D xip for some i and p 2 B or q 2 B. In the latter case, q is also ofthe form q D xip for some i and p 2 B, because B is connected to 1. Thus, inany case, q D xip. Then deg. p/ D k and hence '. p/ D �. p/ by the inductionassumption. From (17.55) and (17.54) we obtain

'.q/ D Xi.'. p// D Xi.�. p// D �.xi�. p// D xi�. p/C g

for some g 2 NL. Similarly, by (17.54), q D xip D xi.�. p/ C h/ for someh 2 NL. Thus we have xi�. p/ D q � xih D '.q/ � g, which implies thatxih 2 A. Therefore, since h 2 NL, it follows from Lemma 17.40 that xih 2 NL.Applying the projection � to the identity q � xih D '.q/ � g we get �.q/ D�.'.q// D '.q/. This completes the induction.

(ii) First we show that

L. pq/ D L.'. pq// for q 2 B; p 2 A: (17.56)


We proceed by induction on deg. p/. For constant p this follows from (i), sinceL.q/ D L.�.q// D L.'.q// for q 2 A. Assume that (17.56) holds for all p 2 Awith deg. p/ D k. Suppose that f 2 A and deg. f / D k C 1. Arguing as in theproof of (i), we can assume without loss of generality that f D xip with p 2 B.Thus we obtain

L. fq/ D L. pxiq/ D L. p'.xiq// D L.'. pxiq// (17.57)

by applying first (i) and then the induction hypothesis. Using the definitions of'.g/ and g.X/ for g 2 RdŒx� and applying (17.55) twice we derive

'. p'.xiq// D p.X/.'.xiq// D P.X/.Xi.q//

D .xip/.X/.q/ D f .X/.q/ D f .X/'.q/ D '. fq/:

Inserting this into (17.57) we get L. pq/ D L.'. pq//: This proves (17.56).Let p; q 2 A. From (i) we obtain L. pq/ D L. p'.q//. Using first this equality,

then (17.56), and finally (17.55) we derive

L. pq/ D L. p'.q// D L.'. p'.q// D L.P.X/'.q// D L.'. pq//

which completes the proof. utNow we define a linear functional QL on RdŒx� by

QL. f / D L.'. f //; f 2 RdŒx�: (17.58)

Lemma 17.42(ii) implies that QL. f / D L. f / for f D pq, where p; q 2 A, and hencefor all f 2 A2. Therefore, QL is an extension of L.

The first statement of the next lemma shows that QL is flat with respect to A.

Lemma 17.43

(i) rankH. QL/ D rankH.L/ D dimD.(ii) NQL D I.NL/ D ker' and VQL D VL:

Proof First we show that

I.NL/ ker' NQL: (17.59)

If p 2 NL, then �. p/ D 0 and hence '. p/ D p.X/1 D 0, so that p 2 ker'. From(17.55) we conclude that ker' is an ideal of RdŒx�. Hence I.NL/ ker'.

Now we verify the second inclusion of (17.59). Let f 2 ker'. Then, by (17.55),for p 2 RdŒx� we obtain

QL. fp/ D L.'. fp// D L. p.X/'. f // D 0;

so that f 2 NQL. This proves (17.59).

17.8 Exercises 441

From (17.54) and Lemma 17.42(i) it follows that RdŒx� D D C I.NL/. Usingthis equality, the corresponding definitions, Proposition 17.17(ii), and (17.59) wederive

dimD D dim.A=NL/ D rankH.L/ � rankH. QL/;rankH. QL/ D dim.RdŒx�=NQL/ � dim.RdŒx�= ker '/

� dim.RdŒx�=I.NL// � dimD:

Hence we have equalities throughout and therefore also in (17.59). This proves (i)and the first assertion of (ii). Since NL and I.NL/ define the same real algebraic setVL, the equality NQL D I.NL/ implies that VQL D VL: utLemma 17.44 If L0 is another linear functional on RdŒx� such that L0 is anextension of L and L0 is flat with respect to A, then L0 D QL.Proof Since L0 is flat with respect to A, NL D NL0 \ A by Proposition 17.33.Obviously, NL0 is an ideal. Hence ker' D J .NL/ NL0 by Lemma 17.43(ii).

Let f 2 RdŒx�. Then '. f / 2 D A and hence '.'. f // D �.'. f // D '. f / byLemma 17.42(i), so that . f � '. f // 2 ker' NL0 . Therefore, L0. f � '. f // D 0.Since L0dD D L, we get L0. f / D L0.'. f // D L.'. f // D QL. f /. Thus, L0 D QL. utLemma 17.45 Suppose that L. p2/ � 0 for p 2 B. Then QL is positive on RdŒx�:

Proof By Proposition 17.34, L is positive on A2. Since QL is flat with respect to A byLemma 17.43(i), the assertion follows by repeated application of Proposition 17.34.Alternatively, one can also argue as follows.

Let f 2 RdŒx�: Then g WD . f�'. f // 2 ker' D NQL as noted in the proof ofLemma 17.44. Hence QL.2'. f /gC g2/ D 0. Since L is positive on A2 and '. f / 2 A,

QL. f 2/ D QL..'. f /C g/2/ D QL.'. f /2/C QL.2'. f /gC g2/ D L.'. f /2/ � 0: ut

Putting the preceding lemmas together we have proved Theorem 17.35. ut

17.8 Exercises

1. Give an example of a truncated K-moment functional and a closed set K suchthat ML;K is not vaguely compact.

2. Let p 2 RdŒx�2n and let Lp 2 HdC1;2n be defined by (17.15).

a. Show that p 2PRŒx�2n if and only if Lp 2P.HdC1;n/2:b. Show that the following are equivalent:

(i) p.x/ � 0 for all x 2 Rd.(ii) Lp.y/ � 0 for all y 2 RdC1.

(iii) Lp.y/ � 0 for all y 2 Sd:


3. Show that for a linear functional L on HdC1;2n the following are equivalent:

(i) There exists a Radon measure on RdC1 such that p 2 L1.RdC1; �/ andL. p/ D R p d� for p 2 HdC1;2n:

(ii) There is a Radon measure on Sd such that L. p/ D R p d� for p 2 HdC1;2n:(iii) L. p/ � 0 for all p 2 HdC1;2n such that p � 0 on Sd:

4. Let L be a linear functional on HdC1;2n given by a k-atomic measure on RdC1.Show that L has a k0-atomic representing measure on Sd, where k0 � k. Findexamples where k0 D k and examples where k0 D k is impossible.

5. Prove that for a linear functional L on RdŒx�2n the following are equivalent:

(i) L is a truncated moment functional.(ii) There exist a number k 2 N and an extension QL of L to a positive linear

functional on RdŒx�2nC2k such that rank HnCk.L/ D rank HnCk�1.L/:

Hint: For instance, combine Theorems 17.15 and 17.29 and Proposition 17.21.6. Let L be a positive functional on A2. Give a second proof for both assertions of

Proposition 17.33 by applying Proposition A.25 to Hankel matrices.7. Show that Proposition 17.27 remains valid for arbitrary sets N. Use this result to

give another proof of the determinacy assertion of Theorem 17.29.8. ([CFM, Lemma 2.5]) Let m1; : : : ;mk 2 R and let x1; : : : ; xk 2 Rd be pairwise

distinct points. Define a functional L on A2 by L. f / D PkjD1mjf .xj/; f 2 A2.

Suppose that k D rankH.L/: Use formula (17.47) to prove that the following areequivalent:

(i) L. f 2/ � 0 for all f 2 A.(ii) m1 � 0; : : : ;mk � 0.

(iii) m1 > 0; : : : ;mk > 0.

17.9 Notes

The multidimensional truncated moment problem was first investigated in the(unpublished) Thesis of J. Matzke [Mt] and independently by R. Curto and L.Fialkow [CF2, CF3]. It is now an active research area, see e.g. [Pu4, La1, CF5,FN1, FN2, BlLs, Bl2, Sm10, F1, DSm1, DSm2]. Multidimensional trigonometrictruncated moment problems and related extension problems are treated in themonograph [BW].

Theorem 17.7 is from J. Stochel [St2], while Theorem 17.13 was provedin [Sm10]. Theorem 17.15 was obtained in [Mt] and independently in [CF5].Theorem 17.29 is contained in [CF2]. Minimal moment functionals have been firststudied in [CFM] where they are called “extremal”.

The flat extension Theorem 17.36 was obtained by R. Curto and L. Fialkow[CF2]. Theorem 17.35 and the proof presented in the text are taken from [LM];see [MS] for a general result that applies to noncommutative �-algebras as well.Another approach to the flat extension theory is given in [Vs3].

17.9 Notes 443

The following result was proved in [CF4] (see [F1] for a correction): Supposethat N D N2

0;n, n 2 N; n � 2, and p 2 RŒx1; x2�, deg. p/ � 2. If L is a linearfunctional on RŒx1; x2�2n satisfying the positivity condition (17.41), the consistencycondition (17.43), and p 2 NL, then L is a truncated moment functional.

Chapter 18Multidimensional Truncated Moment Problems:Basic Concepts and Special Topics

While the preceding chapter dealt with existence questions, this chapter is devoted tofundamental notions and various special topics concerning the structure of solutions.

In Sect. 18.1, we begin the study of the cone of truncated K-moment functionals.We investigate its extreme points, introduce its Carathéodory number and considerextreme points of the solution set.

In Sects. 18.2 and 18.3, we define two real algebraic sets VC.L;K/ and V.L/associated with a truncated K-moment functional resp. a linear functional L onA. These sets are fundamental concepts for truncated moment problems. Theycontain the supports of representing measures. The set V.L/ is called the corevariety of L. If L is a moment functional, then V.L/ coincides with the set ofatoms of representing measures (Theorem 18.21) and L is determinate if and only ifjV.L/j � dim.AdV.L// (Theorem 18.23). We show that a linear functional L ¤ 0

on A is a truncated moment functional if and only if L.e/ > 0 and V.L/ is not empty(Theorem 18.22).

Section 18.4 deals with the maximal mass L.t/ of the one point set ftg amongall representing measures of L. In Sect. 18.5, we construct ordered maximal massrepresentations (Theorem 18.38). In Sect. 18.6, we use evaluation polynomials forthe study of truncated moment problems (Theorems 18.42, 18.43, and 18.46).

In this chapter, N denotes a fixed finite subset of Nd0, A is the linear span of

monomials x˛, ˛ 2 N, and, if not specified otherwise, K is a closed subset of Rd.Further, we retain condition (17.1) from Chap. 17 and assume the following:

There exists an element e 2 A such that e.x/ � 1 for x 2 K: (18.1)

18.1 The Cone of Truncated Moment Functionals

In this section, we use some concepts and results on convex sets from Appendix A.6.The main objects of this chapter are introduced in the following definition.


445

446 18 Multidimensional Truncated Moment Problems: Basic Concepts and Special. . .

Definition 18.1 The moment cone S.A;K/ is the set of truncated K-momentsequences .s˛/˛2N and L.A;K/ is the set of truncated K-moment functionals on A.

Clearly, L.A;K/ is a cone in the dual space A�, S.A;K/ is a cone in RjNj and themap s 7! Ls is a bijection of RjNj and A� which maps S.A;K/ onto L.A;K/.

Set E D AdK and X D K. Then we are in the setup of Section 1.2. Clearly,f 7! f dK maps Pos.A;K/ onto EC. As discussed in Sect. 17.1, L 7! QL is a bijectionof L.A;K/ onto the cone L of moment functionals of E C.KIR/. Here QL isthe linear functional on E defined by QL. f dK/ D L. f /; f 2 A, see (17.6) and Lemma17.4. Repeating verbatim the reasoning from the proof of Proposition 1.27 we obtain

L.A;K/ Pos.A;K/^ D L.A;K/: (18.2)

If L 2 L.A;K/;L ¤ 0, condition (18.1) implies that L.e/ > 0. Thus, L 7! L.e/ is astrictly L.A;K/-positive linear functional on A�. Hence, by (A.24), the convex set

L1.A;K/ WD fL 2 L.A;K/ W L.e/ D 1g (18.3)

is a base of the cone L.A;K/, see Definition A.38. If the set K is compact, there arefurther and stronger results.

Proposition 18.2 Suppose that K is compact.

(i) L.A;K/ D Pos.A;K/^.(ii) S.A;K/ is closed inRjNj and L.A;K/ is closed in the unique norm topology of

the finitedimensional vector space A�:(iii) L1.A;K/ is compact base of the cone L.A;K/ in A�.

Proof

(i) is a restatement of Theorem 17.3.(ii) Since L is closed in E� by Proposition 1.26(ii) and L 7! QL is a linear bijection

of L.A;K/ onto L, so is L.A;K/ in A�. Therefore, since the homeomorphisms 7! Ls of RjNj on A� maps S.A;K/ onto L.A;K/, the cone S.A;K/ is closedin RjNj:

(iii) Let K be a ball in Rd containing K. We equip A with the supremum norm k � kKon K. The norm topology of A� is given by the dual norm of k � kK on A�. LetL 2 L1.A;K/. Then, by Theorem 17.2, L has a k-atomic representing measure� D Pk

jD1 cjıxj , where k � jNj D dimA and cj > 0, xj 2 K. Using condition(18.1) we obtain L.e/ D 1 D P

j cje.xj/ �P

j cj, so that cj � 1 for all j.Therefore,

jL. f /j �kX

jD1cjj f .xj/j �

kXjD1k fkK � jNj k fkK ; f 2 A:

18.1 The Cone of Truncated Moment Functionals 447

Hence L1.A;K/ is bounded in A�. Since the linear functional e 7! L.e/is continuous on A�, L1.A;K/ is obviously closed in A�. Thus, L1.A;K/ iscompact. ut

Example 9.30 shows that for noncompact closed subsets K the cone S.A;K/ isnot closed in general.

Proposition 18.3 Let .P

A2/^ denote the cone of positive functionals on A2.

(i) The cone .P

A2/^ is pointed, that is, .P

A2/^ \ .�.P A2/^/ D f0g:(ii) For each x 2 Rd, the point evaluation lx at x spans an extreme ray of .

PA2/^:

Proof

(i) Let L 2 .PA2/^ \ .�.PA2/^/. Then L. f 2/ � 0 and �L. f 2/ � 0, henceL. f 2/ D 0, for all f 2 A. Therefore, it follows at once from the Cauchy–Schwarz inequality (2.7) that L. fg/ D 0 for f ; g 2 A. Hence L D 0 on A2.

(ii) Let L1;L2 2 .P A2/^. Assume that there are numbers c1 > 0; c2 > 0 such thatlx D c1L1 C c2L2. Set q WD e.x/�1e. Then lx.q/ D 1.

Fix j 2 f1; 2g. Let f ; g; h 2 A. Since Lj is a positive functional on A2, theCauchy–Schwarz inequality (2.7) holds. Using this inequality we derive

0 � cjLj.. f � f .x/q/h/2 � Lj.h2/cjLj. f � f .x/q/2

� Lj.h2/Œc1L1. f � f .x/q/2 C c2L2. f � f .x/q/2� D Lj.h

2/lx. f � f .x/q/2 D 0:

Here the last two equalities hold by the relations lx D c1L1Cc2L2 and lx.q/ D 1and by the definition of the evaluation functional lx. Hence, since cj > 0, wehave Lj.. f � f .x/q/h/ D 0, so that

Lj.fh/ D f .x/Lj.qh/ for f ; h 2 A: (18.4)

Put aj WD Lj.q2/: Setting h D g � g.x/q in (18.4) we obtain

Lj.fg/� g.x/Lj.fq/ D f .x/Lj.gq/� f .x/g.x/aj: (18.5)

For h D q, (18.4) yields Lj.fq/ D f .x/aj. Replacing f by g, we get Lj.gq/ Dg.x/aj. Inserting these two facts into (18.5) it follows that Lj.fg/ D f .x/g.x/aj Dajlx.fg/ for arbitrary f ; g 2 A. Therefore, Lj D ajlx on A2, that is, Lj belongs tothe ray spanned by lx. This proves that lx spans an extreme ray. ut

Example 18.4 (An extreme ray of .P

A2/^ which is not spanned by a pointevaluation) Let A2 D R2Œ x �6 be the polynomials in two variables of degree atmost 6.

We denote by Exr the set of elements of all extreme rays of the cone .P

A2/^.By Lemma 18.3(i), .

PA2/^ is pointed. The cone .

PA2/^ is obviously closed in

the norm topology of the dual of R2Œ x �6. Thus, Proposition A.37 applies and showsthat each element of .

PA2/^ is a finite sum of elements of Exr.


On the other hand, by Proposition 13.5, there exists a positive linear functionalL2 on R2Œ x � such that L2. pc/ < 0 for the Motzkin polynomial pc. The restrictionL D L2dR2Œ x �6 is in .

PA2/^, but L is not a truncated moment functional, since

pc � 0 on R2. Hence L is not a positive combination of point evaluations. SinceL 2 .PA2/^ is a sum of elements of Exr, we conclude that there exists an extremeray of .

PA2/^ that is not spanned by a point evaluation. As shown in [Bl1, Theorem

1.6], the Hankel matrix of each nonzero functional contained in such an extreme rayhas rank 7. ı

Example 18.4 shows that extreme rays of the cone .P

A2/^ are not necessarilyspanned by point evaluations. But all extreme rays of the smaller cone of truncatedK-moment functionals do, as shown in the next proposition.

Proposition 18.5 For a truncated K-moment functional L ¤ 0 on A2 the followingstatements are equivalent:

(i) L spans an extreme ray of the cone L.A2;K/.(ii) L has a 1-atomic representing measure.

(iii) There are a point x 2 K and a number c > 0 such that L D clx.(iv) rankH.L/ D 1.Proof

(i)!(ii) Let � D PkjD1mjıxj 2 ML with the smallest number k of atoms.

Thus, L D PkjD1mjlxj . Assume to the contrary that (ii) is not true. Then we have

k � 2 and L D m1L1 C L2, where L1 WD lx1 and L2 WDPkjD2mjlxj . Clearly, m1 > 0

and L2 ¤ 0. Since L spans an extreme ray, Lj D ajL for some aj > 0; j D 1; 2. Thus,lx1 D L1 D a1a�1

2 L2. Hence � has less than k atoms, which is a contradiction.(ii)$(iii) is obvious.(iii)!(i) By Proposition 18.3, lx generates an extreme ray of the cone of positive

functionals and hence also of the subcone L.A;K/.(ii)!(iv) Let � D cıx 2 ML. Recall that s.x/ � sN.x/ denotes the vector

.x˛/˛2N 2 RjNj. By formula (19.29), we have H.L/ D c s.x/s.x/T , which impliesthat rankH.L/ � 1. Since L ¤ 0, H.L/ ¤ 0: Hence rankH.L/ D 1.

(iv)!(iii) Since H.L/ � 0 has rank one, there is a nonzero vector � 2 RjNj suchthat H.L/ D ��T . Let� DPk

jD1mjıxj be a representing measure of L, where mj > 0

for all j. Then, H.L/ DPkjD1mj s.xj/s.xj/T by (19.29). For 2 RjNj we obtain

TH.L/ DkX

jD1mjjh; s.xj/ij2 D jh; �ij2; (18.6)

where h�; �i is the canonical scalar product of RjNj: From (18.6) and mj > 0 it followsthat ? � implies ? s.xj/: Hence all vectors s.xj/ are multiples of �: Therefore,since � ¤ 0, there exists an i such that �, hence all s.xj/ are multiplies of s.xi/. ThenH.L/ D c s.xi/s.xi/T with c > 0.

18.1 The Cone of Truncated Moment Functionals 449

We prove that L D clxi : Let f DP˛2N f˛x˛ 2 A and set Ef D . f˛/T 2 RjNj: Then

hEf ; sxii DP

˛ f˛.xi/˛ D f .xi/: Therefore, using (17.33) for f ; g 2 A we obtain

L. fg/ D Ef TH.L/Eg D c hEf ; s.xi/i hEg; s.xi/i D cf .xi/g.xi/ D clxi.fg/:

Since A2 is spanned by products fg with f ; g 2 A, we get L D clxi : utFor general vector spaces not of the form A2, point evaluations do not necessarily

span extreme rays, as shown by the following simple example.

Example 18.6 (A point evaluation on A which does not span an extreme ray) Letd D 1;N D f0; 1g, so that A D Lin f1; xg. Then l0 D 1

2.l1 C l�1/ on A, but l1 ¤ al0

on A for all a � 0. Hence l0 does not span an extreme ray of the cone L.A;R/. ıThe next definition contains two other important notions.

Definition 18.7 Let s 2 S.A;K/ and L 2 L.A;K/. The Carathéodory numberC.s/ resp. C.L/ is the smallest number k 2 N0 such that s resp. L has a k-atomicrepresenting measure � 2 MC.K/.

The Carathéodory number C.A;K/ is the largest C.s/, where s 2 S.A;K/.

Thus, C.A;K/ the smallest number n 2 N such that each sequence s 2 S.A;K/has a k-atomic representing measure, where k � n. Obviously, C.s/ D C.Ls/. Bydefinition we have C.L/ D 0 for L D 0.

Recall from Corollary 1.25 that each moment sequence s 2 S.A;K/ has a k-atomic representing measure, where k � jNj. Therefore, the Carathéodory numbersC.s/, C.L/; and C.A;K/ are well-defined and satisfy

C.A;K/ D sup fC.s/ W s 2 S.A;K/ g D sup fC.L/ W L 2 L.A;K/ g � jNj D dim A:

Remark 18.8 Let P be a cone in a finite-dimensional real vector space E ¤ f0g:Let Exr.P/ denote the set of elements of extreme rays of P. We assume that eachelement of P is a finite sum of elements of Exr.P/. (By Proposition A.37, this holdsif P is pointed and closed. It is also true for the cones S.A2;K/ and L.A2;K/ byTheorem 17.2 and Proposition 18.5, even though these cones are not necessarilyclosed.)

For c 2 P, let C.c/ denote the smallest number k such that c is a sum of k elementsfrom Exr.P/. By Carathéodory’s theorem, C.c/ � dim E. The largest number C.c/for c 2 P is called the Carathéodory number C.P/ of the cone P.

Now let P D S.A2;K/ and let E be the vector space spanned by S.A2;K/. ThenExr.P/ D f�lx W � � 0; x 2 Kg by Proposition 18.5. Thus, in this case theCarathéodory number C.P/ of the cone P is just the Carathéodory number C.A;K/according to Definition 18.7. ıExample 18.9 (Moment cone and Carathéodory number in dimension one onRŒx�m)Let d D 1;m 2 N; N D f0; 1; : : : ;mg, and K WD Œa; b�; where a; b 2 R; a < b:


Then A D RŒx�m and S.A;K/ is just the moment cone SmC1 from Definition 10.4.For the Carathéodory number we have

C.A;K/ D nC 1 if m D 2n or m D 2nC 1; n 2 N:

That is, if bm2c denotes the largest integer k satisfying k � m

2, we can write

C.A;K/ Djm2

kC 1: (18.7)

We prove formula (18.7) in the even case m D 2n; the odd case is similar. For thiswe use some notions and facts from Sects. 10.2 and 10.3. Let s ¤ 0 be a sequenceof the moment cone SmC1 and choose a representing measure � D Pk

jD1mjıxj ofs such that ind .�/ D ind .s/. Recall that for the definition of ind .�/ atoms in.a; b/ are counted twice, while possible atoms a or b are counted only once. If s isa boundary point of SmC1, then ind .�/ � m by Theorem 10.7. Hence � has atmost n C 1 atoms. If s is an interior point of SmC1, it has a principal representingmeasure �C with nC 1 atoms by (10.14). Since ind .s/ � mC 1 D 2nC 1 (againby Theorem 10.7), s has no representing measure with fewer atoms than nC1. Thisproves that C.A;K/ D nC 1 D bm

2c C 1: ı

Determining the Carathéodory number in the multivariate case is a difficultproblem and only a few results are known, see Theorem 18.43(ii) and Sect. 19.1below.

Our next result characterizes the extreme points of the set ML;K.

Proposition 18.10 Suppose that L ¤ 0 is a truncated K-moment functional on Aand � 2 ML;K. Let J� denote the canonical embedding of A into L1R.R

d; �/. Thefollowing statements are equivalent:

(i) � is an extreme point of the set ML;K.(ii) J�.A/ D L1R.R

d; �/.(iii) � is k-atomic with atoms t1; : : : ; tk 2 K and there exist elements p1; : : : ; pk 2 A

such that pj.ti/ D ıij for i; j D 1; : : : ; k.Proof

(i)$(ii) By Proposition 1.21,� is an extreme point of the set ML;K if and only ifthe image of A is dense in L1R.R

d; �/. Since A is finite-dimensional, the latter holdsif and only if J�.A/ D L1R.R

d; �/. Thus (i) and (ii) are equivalent.(iii)!(ii) is obvious.(ii)!(iii) Because N is finite, J�.A/ D L1R.R

d; �/ (by (ii)) is finite-dimensional.Hence the measure � is k-atomic, where k D dim L1R.R

d; �/: Let � DPklD1 mlıtl

with t1; : : : ; tk 2 K. Since J�.A/ D L1R.Rd; �/, the characteristic function �tj of

each singleton ftjg, j D 1; : : : ; k, is in the image J�.A/. Thus �tj D J�. pj/ for somepolynomial pj 2 A: Then pj.ti/ D ıij for i; j D 1; : : : ; k. ut

18.2 Inner Moment Functionals and Boundary Moment Functionals 451

Condition (iii) in Proposition 18.10 suggests the following definition.

Definition 18.11 We say that points t1; : : : ; tk of K satisfy the separation property.SP/A if there exist elements p1; : : : ; pk 2 A such that pj.ti/ D ıij; i; j D 1; : : : ; k.

Proposition 18.10 (i)$(iii) says for each truncated K-moment functional L ¤ 0the extreme points of ML;K are precisely those k-atomic measures in ML;K forwhich all atoms are in K and satisfy property .SP/A.

We close this section with simple but useful determinacy criterion.

Proposition 18.12 Let p 2 Pos.A;K/. Suppose that Z. p/ \ K D ft1; : : : ; tkg andt1; : : : ; tk 2 K satisfy .SP/A. Let � D Pk

jD1mjıtj ; where mj � 0 for all j. If L is atruncated K-moment functional and � 2ML;K, then the set ML;K D f�g, that is,L is a determinate truncated K-moment functional.

Proof Let � be another measure from ML;K. Since p 2 Pos.A;K/ andZKp d� D L. p/ D

Zp d� D

Xjmjp.tj/ D 0;

Proposition 1.23 implies that supp � Z. p/\ K D ft1; : : : ; tkg. Hence � is of theform � DPk

jD1 njıtj for some numbers nj � 0:Now fix i 2 f1; : : : ; kg. Since t1; : : : ; tk satisfy .SP/A, there exists a q 2 A such

that q.tj/ D ıij for j D 1; : : : ; k. Then L.q/ D R q d� D mi and L.q/ D R q d� D ni,which implies that mi D ni. This proves that � D �, that is, L is determinate. ut

18.2 Inner Moment Functionals and Boundary MomentFunctionals

Throughout this section, L is a truncatedK-moment functional on A:

Definition 18.13

NC.L;K/ WD fp 2 Pos.A;K/ W L. p/ D 0 g;VC.L;K/ WD fx 2 Rd W p.x/ D 0 for p 2 NC.L;K/g:

Clearly, NC.L;K/ is a subcone of the cone Pos.A;K/ and VC.L;K/ is a realalgebraic set in Rd. In the case K D Rd we write

NC.L/ WD NC.L;Rd/; VC.L/ WD VC.L;Rd/: (18.8)

Remark 18.14 We compare NC.L;K/ and VC.L;K/ with their counterparts NL

and VL from Definition 17.16. For this we suppose that the linear functional L isdefined on A2. Then the definition of NC.L;K/ refers to elements from A2, while


NL contains only polynomials from A by (17.29). Obviously, if p 2 NL, thenp2 2 NC.L;K/. Therefore, for any closed set K we have

VC.L;K/ VL; (18.9)

but in general VC.L;K/ ¤ VL. In Sect. 19.6 we develop examples for which VL ¤VC.L/. ı

The next proposition is the counterpart of Proposition 17.18 in the present setting.

Proposition 18.15

(i) For each representing measure � 2ML;K we have supp� VC.L;K/ \K.(ii) Let p 2 NC.L;K/ and q 2 A be such that p q 2 A. Then L. p q/ D 0. If in

addition q 2 Pos.A;K/, then p q 2 NC.L;K/.

Proof

(i) follows at once from Proposition 1.23.(ii) The proof is almost verbatim the same as the proof of Proposition 17.18. Take

a measure � 2ML;K: Using that supp� VC.L;K/ \K by (i) and p.x/ D 0on VC.L;K/ by definition, we obtain

L. p q/ DZK. p q/.x/ d�.x/ D

ZVC.L;K/\K

p.x/q.x/ d�.x/ D 0: (18.10)

If q is in Pos.A;K/, so is p q and therefore p q 2 NC.L;K/. utThe set NC.L;K/ is closely related to the set NC.L/ from Definition 1.39. Since

L is a truncated K-moment functional, there is a moment functional QL on E WD AdKdefined by QL. f dK/ WD L. f /; f 2 A: Clearly, we have f 2 NC.L;K/ if and only iff dK 2 NC. QL/, that is, NC. QL/ D NC.L;K/dK: Therefore, Proposition 1.42 may berestated as the following.

Proposition 18.16

(i) Let p 2 NC.L;K/; pdK ¤ 0. Then 'p.L0/ D L0. p/; L0 2 A�, defines asupporting functional 'p of the coneL.A;K/ at L. Each supporting functionalof L.A;K/ at L is of this form.

(ii) L is a boundary point of the cone L.A;K/ if and only if NC.L;K/dK ¤ f0g:(iii) L is an inner point of the cone L.A;K/ if and only if NC.L;K/dK D f0g:

Note that if p 2 A vanishes on K, then obviously p 2 NC.L;K/, but 'p � 0, so'p is not a supporting hyperplane of L.A;K/ at L. That is, the requirement pdK ¤ 0cannot be omitted in Proposition 18.16(i).

Since L 2 Pos.A;K/^, NC.L;K/ is an exposed face of the cone Pos.A;K/. Thenonempty exposed faces of the cone L.A;K/ are exactly the sets

Fp WD fL0 2 L.A;K/ W 'p.L0/ � L0. p/ D 0g; where p 2 Pos.A;K/:

18.3 The Core Variety of a Linear Functional 453

Example 18.17 (Finitely atomic representing measure) Let � D PkjD1 cjıxj be a

representing measure of L, where xj 2 K and cj > 0 for all j. Clearly, then we have

NC.L;K/ D fp 2 Pos.A;K/ W p.x1/ D � � � D p.xk/ D 0 g: (18.11)

Whether or not VC.L;K/ is “small” or “large” depends very sensitively on the sizeof A and the number and the location of atoms xj. Roughly speaking, if sufficientlymany atoms are well distributed in a real algebraic set, this set is contained inVC.L;K/. We illustrate this by the following very simple example. Similar resultsfor R2Œ x �4 and R2Œ x �6 will be given by Propositions 19.35 and 19.37 below. ıExample 18.18 Let d D 2;N D f.n1; n2/ 2 N2

0 W n1 C n2 � 2g;K D R2. Then A isthe space of polynomials in two variables of degree at most 2. Let L be a truncatedmoment functional and � DPk

jD1 cjıxj a k-atomic representing measure of L.Clearly, each element of Pos.A;K/ is a sum of squares of linear polynomials. If

k D 1, then VC.L/ D fx1g. Suppose that k � 2. Then, if all points xj are on a line,then VC.L/ is this line, and if the points xj are not on a line, then VC.L/ D R2. ı

18.3 The Core Variety of a Linear Functional

In this section, L is a linear functional on A and K D Rd, that is, we consider thetruncated moment problem on Rd. Our aim is to study the core variety V.L/ of L.

We define inductively linear subspaces Nk.L/, k 2 N; of A and real algebraic setsVj.L/, j 2 N0; by V0.L/ D Rd,

Nk.L/ WD fp 2 Pos.A;Vk�1.L// W L. p/ D 0 g; (18.12)

Vj.L/ WD ft 2 Rd W p.t/ D 0 for p 2 Nj.L/g: (18.13)

If Vk.L/ is empty for some k, we set Vj.L/ D Vk.L/ D ; for all j � k; j 2 N.For k D 1 these notions coincide with those defined in Eq. (18.8), that is,

N1.L/ D NC.L/ � NC.L;Rd/ and V1.L/ D VC.L/ � VC.L;Rd/:

Definition 18.19 The core variety V.L/ of the linear functional L on A is

V.L/ WD1\jD0

Vj.L/: (18.14)

Since V.L/ is the zero set of real polynomials, V.L/ is a real algebraic set in Rd.The preceding definitions resemble the corresponding definitions in Sect. 1.2.5.

More precisely, if we set X D Rd, E D A C.RdIR/ and L is a moment functionalon E, then formulas (1.22), (1.23), (1.24) coincide with (18.12), (18.13), (18.14),respectively, that is, we have Nk.L/ D Nk.L/, Vj.L/ D Vj.L/, V.L/ D V.L/.


Note that L is now an arbitrary linear functional on A, while it was a momentfunctional in Subsection 1.2.5.

Proposition 18.20

(i) Nk.L/ NkC1.L/ and Vk.L/ Vk�1.L/ for k 2 N.(ii) If L is a truncated moment functional on A and � 2ML, then supp� V.L/:

(iii) There exists a k 2 N0 such that Vk.L/ D VkC1.L/; for any such k we have

V.L/ D Vk.L/ D Vn.L/; n � k: (18.15)

Proof The proofs of (i) and (ii) are verbatim the same as the proofs of assertions (i)and (ii) of Proposition 1.48.

(iii) Define an ideal Ij of RdŒ x � by Ij D fp 2 RdŒ x � W p.x/ D 0 for x 2 Vj.L/gfor j 2 N0: Since VjC1.L/ Vj.L/ by (i), we have Ij IjC1. Thus,

I0 I1 � � � Ij IjC1 � � �

is an ascending chain of ideals in RdŒ x �. Since RdŒ x � is a Noetherian ring(see e.g. [CLO, p. 77]), there exists a k 2 N0 such that Ik D In for n � k: LetUj denote the real algebraic set defined by Ij. Since Ij is the vanishing idealof Vj.L/, it follows that Uj D Vj.L/. Clearly, Ik D In implies Uk D Un, so thatVk.L/ D Vn.L/ for n � k.

Assume that Vk.L/ D VkC1.L/ for some k 2 N0. Then NkC1.L/ DNkC2.L/ by (18.12) and hence VkC1.L/ D VkC2.L/ by (18.13). Proceedingby induction we get Vn.L/ D VnC1.L/ for n � k, which implies (18.15). ut

For a truncated moment functional L on A we define the set of possible atoms:

W.L/ WD fx 2 Rd W �.fxg/ > 0 for some � 2MLg: (18.16)

The importance of the core variety stems from the following three theorems.The main assertion of the first theorem is only a restatement of Theorem 1.49. It

says that the set W.L/ of atoms is equal to the real algebraic set V.L/; in particular,W.L/ is a closed subset of Rd.

Theorem 18.21 Let L be a truncated moment functional on A. Then

W.L/ D V.L/: (18.17)

Each representing measure � of L is supported on V.L/. For each point x 2 V.L/there is a finitely atomic representing measure � of L which has x as an atom.

Proof Theorem 1.49 applies with E D A; X D Rd and yields the equality (18.17),since V.L/ D V.L/ and W.L/ D W.L/. The other assertions follow fromProposition 18.20(ii) and Corollary 1.25. ut

18.3 The Core Variety of a Linear Functional 455

The next results provide existence and determinacy criteria in terms of the corevariety V.L/; they are based on Theorems 1.30 and 1.36.

Recall that by condition (18.1) there exists an e 2 A such that e.x/ � 1 onK D Rd.

Theorem 18.22 A linear functional L ¤ 0 on A is a truncated moment functionalif and only if L.e/ � 0 and the core variety V.L/ is not empty.Proof In this proof we abbreviate Nj D Nj.L/, Vj D Vj.L/, and V D V.L/.

First suppose L is a truncated moment functional. Since each representingmeasure is supported on V by Theorem 18.21 and L ¤ 0, we have V ¤ ; andL.e/ > 0.

Now we prove the converse implication. If L.e/ D 0, then e 2 N1, so V is empty,which contradicts the assumption. Therefore, L.e/ > 0. By Proposition 18.20(iii),there is a number k 2 N0 such that V D Vk D VkC1.

Let f 2 Pos.A;V/. We prove that L. f / � 0. Assume to the contrary thatL. f / < 0. Set q WD f � L.e/�1L. f /e 2 A. Since L.e/ > 0, L. f / < 0, ande.x/ � 1 on Rd, we have q.x/ > 0 on V D Vk and L.q/ D 0. Hence q 2 NkC1 andV D VkC1 Z.q/ D ;. This is a contradiction, since V is not empty. Thus we haveshown that L. f / � 0.

Therefore, by Lemma 17.4, condition (17.5) holds. Hence there exists a well-defined linear functional QL on E D AdV given by QL. f d/ WD L. f /; f 2 A:

We prove that QL is strictly EC–positive. Let f 2 A be such that f dV 2 EC andf dV ¤ 0. Then f 2 Pos.A;V/ and hence L. f / � 0 as shown in the paragraph beforelast. Since f dV ¤ 0 and V is not empty, there exists an x0 2 V such that f .x0/ > 0.If L. f / were zero, then f � 0 on Vk implies f 2 NkC1, so V D VkC1 Z. f /: Thisis impossible, because x0 2 V and f .x0/ > 0: Thus QL is strictly EC–positive.

Therefore, by Theorem 1.30, QL is a moment functional on E and so is L on A. utTheorem 18.23 For any truncated moment functional L on A the following areequivalent:

(i) L is not determinate.(ii) jV.L/j > dim.AdV.L//:

(iii) There is a representing measure � of L such that jsupp�j > dim.AdV.L//:In particular, L is not determinate if jV.L/j > dimA D jNj or if L has arepresenting measure � such that jsupp�j > dimA D jNj.Proof Since W.L/ D V.L/ by Theorem 18.21, the assertions are only restatementsof Theorem 1.36 and Corollary 1.37, applied with E D A and X D Rd. ut

Suppose that L is a truncated moment functional on A. Let k be the smallestinteger for which Vk.L/ D VkC1.L/. Then, by Proposition 18.20 and (18.17),

W.L/ D V.L/ D Vn.L/ D Vk.L/ ¤ Vj.L/ ¤ V1.L/ (18.18)


for n � k; 1 < j < k: In general, it is difficult to determine the set V.L/ D W.L/.It would be important to have useful criteria ensuring that V.L/ D V1.L/ � VC.L/;see also Theorem 1.45.

We close this section with two illustrating examples.

Example 18.24 (V.L/ ¤ ; and L is not a truncated moment functional) Let d D 1and N D f0; 2g. Then A D faC bx2 W a; b 2 Rg, so condition (18.1) is satisfied withe.x/ D 1. Define a linear functional L on A by L.aC bx2/ D �a. Clearly, L is not atruncated moment functional, because L.1/ D �1.

Then Pos.A;R/DfaCbx2 W a � 0; b � 0g, so N1.L/ D RC �x2 and V1.L/ D f0g.Hence Pos.A;V1.L// D faCbx2 W a � 0; b 2 Rg, N2.L/ D R �x2, and V2.L/ D f0g.Therefore, V.L/ D f0g ¤ ;: ıExample 18.25 (A truncated moment functional with V.L/ D V2.L/ ¤ V1.L/) Letd D 1 and N D f0; 2; 4; 5; 6; 7; 8g. Then A D Lin f1; x2; x4; x5; x6; x7; x8g. We fix areal number ˛ > 1 and define � D ı�1 C ı1 C ı˛: For the corresponding momentfunctional L � L� D l�1 C l1 C l˛ on A we shall show that

W.L/ D V.L/ D V2.L/ D f1;�1; ˛g f1;�1; ˛;�˛g D V1.L/: (18.19)

Let us prove (18.19). Put p.x/ WD .x2 � 1/2.x2 � ˛2/2. Clearly, p 2 A, L. p/ D 0

and p 2 Pos.R/, so that p 2 N1.L/. Conversely, let f 2 N1.L/. Then L. f / D 0

implies that f .˙1/ D f .˛/ D 0. Since f 2 Pos.R/, the zeros 1;�1; ˛ have evenmultiplicities. Hence f .x/ D .x�1/2.xC1/2.x�˛/2.ax2CbxCc/ with a; b; c 2 R.The coefficients of x and x3 of f are ˛2b�2˛c and 4˛cC.1�2˛2/b�2˛a. Since xand x3 are not in A, these coefficients have to be zero. This yields b D 2˛a; c D ˛2a.Therefore, ax2 C bxC c D a.xC ˛/2. Clearly, a � 0; so that f D ap 2 RC�p. Thuswe have shown that N1.L/ D RC�p. Hence V1.L/ D Z. p/ D f1;�1; ˛;�˛g:

Now we set q.x/ D x4.x2 � 1/.˛ � x/. Then q 2 A. Since q.˙1/ D q.˛/ D 0

and q.�˛/ D ˛4.˛2 � 1/2˛ > 0; we have q 2 Pos.A;V1.L// and L.q/ D 0. Thus,q 2 N2.L/ and hence V2.L/ V1.L/ \ Z.q/ D f1;�1; ˛g:

Since 1;�1; ˛ are atoms of �, f1;�1; ˛g W.L/. Now (18.19) follows from(18.18) and the preceding. Note that the moment functional L is determinate. ı

18.4 Maximal Masses

In this section, we suppose that L ¤ 0 is a truncated K-moment functional on A.The next definition contains the main notions that will be studied in this section.

Definition 18.26

L.x/ WD sup f�.fxg/ W � 2ML;Kg ; x 2 K; (18.20)

W.L;K/ WD fx 2 K W �.fxg/ > 0 for some � 2ML;K g:

18.4 Maximal Masses 457

That is, W.L;K/ is the set of atoms and L.x/ is the supremum of point massesat x of all solutions of the truncated K-moment problem for L. Note that W.L;Rd/

coincides with the set W.L/ defined by (18.16).Since L always has an atomic representing measure by Theorem 17.2 and L ¤ 0

by assumption, W.L;K/ is not empty. It is obvious that

W.L;K/ D fx 2 K W L.x/ > 0g:

By the inequality (18.25) below, L.x/ <1 for all x2K. From Proposition 18.15(i),

W.L;K/ VC.L;K/ \K: (18.21)

In general, W.L;K/ ¤ VC.L;K/\K, as shown by Example 1.41 with K D R2.It is an interesting problem to determine when we have equality in (18.21).

Further, we define another important nonnegative number L.x/ by

L.x/ D inff L. p/ W p 2 Pos.A;K/; p.x/ D 1g (18.22)

D inf

�L. p/

p.x/W p 2 Pos.A;K/

�; (18.23)

where we set c0WD C1 for c � 0: The equality in (18.23) is obvious. The

polynomial px WD e.x/�1e is in Pos.A;K/ and satisfies px.x/ D 1; so thecorresponding set in (18.22) is not empty and L.x/ is well defined.

Note that both quantities L.x/ and L.x/ depend on the set K as well; forsimplicity we have suppressed this dependence.

The number L.x/ is defined by a conic optimization problem (A.25) for thecone C D Pos.A;K/ in A and functionals L1 D L and L0 D lx, see Appendix A.6.The corresponding dual problem (A.26) is

L.x/� WD sup fc 2 R W .L � clx/ 2 Pos.A;K/^ g: (18.24)

A standard fact of the duality theory of conic optimization (see e.g. [BN]) states thatif L is an interior point of the dual cone C^, then the infimum in (A.25) is attained.Proposition 18.28(iii) below is the corresponding result in the present context.

Proposition 18.27 Let x 2 K. Then

L.x/ D cL.x/ WD sup fc 2 RC W .L � clx/ 2 L.A;K/ g � L.x/ < C1 (18.25)

and the functional L � L.x/lx belongs to the boundary of the cone L.A;K/:If K is compact, then we have L.x/ D L.x/, the supremum in Eq. (18.20) is a

maximum, and L � L.x/lx is a truncated K-moment functional.


Proof In this proof, we abbreviate L WD L.A;K/:Let c 2 RC be such that QL WD .L�clx/ 2 L. Such c exists; for instance, c D 0. IfQ� is a representing measure of QL, then� D Q�Ccıx is a positive measure representingL. Hence c � �.fxg/ � L.x/. Taking the supremum over c yields cL.x/ � L.x/.

Let us assume that cL.x/ < L.x/. By the definition of L.x/, there exist a numberc 2 .cL.x/; L.x// and a representing measure � of s such that �.fxg/ D c: ThenQ� WD � � c � ıx is a positive (!) Radon measure representing QL D L � clx. HenceQL 2 L; so that c � cL.x/, which is a contradiction. This proves that cL.x/ � L.x/:

Combining the preceding two paragraphs, we have shown that cL.x/ D L.x/.Let � 2ML;K. For any p 2 Pos.A;K/; p.x/ D 1; we have

L. p/ DZ

p.y/ d�.y/ � �.fxg/p.x/ D �.fxg/:

Taking the infimum over such p and the supremum over � 2 ML;K we obtain L.x/ � L.x/. This completes the proof of (18.25).

Set L0 WD L � L.x/lx D L � cL.x/lx: From the definition of cL.x/ in (18.25) itfollows at once that the functional L0 belongs to the boundary of L.A;K/:

For the rest of this proof we assume that K is compact.Since K is compact, we have L.A;K/ D Pos.A;K/^ by Proposition 18.2(i).

Hence cL.x/ D L.x/� by (18.24). Lemma A.40 yields L.x/ D L.x/�. Therefore, L.x/ D cL.x/ D L.x/: Further, L.A;K/ is closed in A� by Proposition 18.2(ii).Therefore, the boundary functional L0 belongs to the cone L.A;K/.

By the definition of L.x/; there is a sequence .�n/n2N of ML;K such thatlimn �n.fxg/ D L.x/. By Proposition 17.3(ii), ML;K is compact in the vaguetopology. Hence there exists a subnet .�i/i2J of the sequence .�n/n2N whichconverges vaguely to some measure � 2ML;K. Then

L.x/ D limn�n.fxg/ D lim

i�i.fxg/ � �.fxg/;

where the last inequality holds by Proposition A.5, applied to K D fxg. Bydefinition, �.fxg/ � L.x/. Thus, L.x/ D �.fxg/. This shows that the supremumin (18.20) is attained at �. utProposition 18.28 Suppose that x 2 K.

(i) If L.x/ > 0, then x 2 VC.L;K/:(ii) Suppose that K is compact and assume that the infimum in (18.23), or

equivalently in (18.22), is a minimum. Then x 2 VC.L;K/ implies thatL.x/ > 0.

(iii) If L is a relatively interior point of the cone L.A;K/, or equivalently, L. f / > 0for all f 2 Pos.A;K/, f dK ¤ 0, then the infimum in (18.23) is a minimum.

Proof

(i) Suppose that L.x/ > 0 and assume to the contrary that x … VC.L;K/. Thenthere exists a p0 2 NC.L;K/ such that p0.x/ ¤ 0. Upon scaling we can assumethat p0.x/ D 1. Since L. p0/ D 0 by p0 2 NC.L;K/, then L.x/ D 0 by


(18.22). Hence L.x/ D 0, because L.x/ � L.x/ � 0 by (18.25). This is acontradiction.

(ii) Let x 2 VC.L;K/. Let us assume that the infimum in (18.22) is attained atp0 2 Pos.A;K/, that is, L. p0/ D L.x/. If L. p0/were zero, then we would havep0 2 NC.L;K/ and hence p0.x/ D 0, since x 2 VC.L;K/. This contradictsp0.x/ D 1 by (18.22). Thus L. p0/ > 0. Since K is compact, L.x/ D L.x/ byProposition 18.27. Therefore, L.x/ D L.x/ D L. p0/ > 0.

(iii) We will derive the assertion from Theorem 1.30. Let E D AdK, X D K, Qf Df dK: Recall that L 7! QL is a bijection of L.A;K/ onto the coneL of the momentfunctional on E; where QL is defined by (17.6), that is, QL.Qf / D L. f /; f 2 A: Thesecond assumption on L means that QL is strictly EC-positive. By Lemma 1.29,this holds if and only QL is an inner point of L, or equivalently, L is a relativelyinterior point of L.A;K/: Then, by Theorem 1.30(iii), there exists Qf0 2 EC,Qf0.x/ D 1; such that QL. Qf0/ D inf f QL.Qf / W Qf 2 EC; Qf .x/ D 1g: Since QL.Qf / D L. f /and Qf 2 EC if and only if f 2 Pos.A;K/, we conclude that f0 2 A attains theinfimum in (18.23). ut

The following simple example shows that the supremum in (18.20) is notnecessarily attained if the set K is not compact.

Example 18.29 Let K D R and N D f0; 1; 2g. Define L. p/ D 12. p.1/C p.�1// for

p 2 A D RŒx�2: For c 2 Œ0; 1/; we set �c WD cı0 C 12.1 � c/.ıy.c/ C ı�y.c//, where

y.c/ D .1 � c/�1=2: Then �c is a representing measure of L and �c.f0g/ D c.One verifies that L.0/ D cL.0/ D L.0/ D 1. While the infimum for L.0/

in (18.22) is attained at p D 1, the suprema for L.0/ in (18.20) and cL.0/ in (18.25)are not attained. That is, there is no measure � 2 ML such that �.f0g/ D 1 andL0 WD L�L.0/l0 ¤ 0 is not a truncated moment functional on A, since L0.1/ D 0. ıDefinition 18.30 Let � D Pk

jD1 mjıxj be a measure in ML;K, where k 2 N andxj 2 K for j D 1; : : : ; k, and let i 2 f1; : : : ; kg. We shall say that

� � has maximal mass at xi if mi D L.xi/;� � is a maximal mass measure for the functional L, or briefly, � is maximal mass,

if mj D L.xj/ for all j D 1; : : : ; k.

Recall that a measure � D PkjD1 mjıxj is called k-atomic if mj > 0 for all j and

xi ¤ xj for all i ¤ j.The following proposition and Theorem 18.33 below deal with the question of

when masses of atoms of atomic representing measures are maximal masses.

Proposition 18.31 Suppose that � D PkjD1 mjıxj is a representing measure of L,

where k 2 N and mj > 0 and xj 2 K for all j. Let i 2 f1; : : : ; kg. Then:(i) If mi D L.xi/, then mi D L.xi/, that is, � has maximal mass at xi.

(ii) mi D L.xi/ if and only there exists a sequence . pn/n2N with pn 2 Pos.A;K/such that pn.xi/ D 1 and limn pn.xj/ D 0 for all j ¤ i.

(iii) If there exists a p 2 Pos.A;K/ such that p.xi/ D 1 and p.xj/ D 0 for j ¤ i,then mi D L.xi/ D L.xi/.


(iv) Assume that the infimum (18.22) for x D xi is a minimum. If mi D L.xi/; thenthere exists a p 2 Pos.A;K/ such that p.xi/ D 1 and p.xj/ D 0 for j ¤ i.

(v) If mj D L.xj/ for all j D 1; : : : ; k, then k � dimA and � is k-atomic.

Proof By the definition of L.xi/ and (18.25) we always have

mi � L.xi/ � L.xi/: (18.26)

(i) follows at once from (18.26).(ii) First suppose that there exists a sequence . pn/n2N as stated above. Since

pn.xi/ D 1, we have mi � L.xi/ � L. pn/ by (18.23). Combined with theequality

limn

L. pn/ DkX

jD1mj�

limn

pn.xj/� D

kXjD1

mjıji D mi

we conclude that mi D L.xi/.Conversely, suppose that mi D L.xi/. By the definition of L.xi/, there

exists a sequence . pn/n2N of elements pn 2 Pos.A;K/ such that pn.xi/ D 1

and L.xi/ D limn L. pn/. Clearly, 0 � pn.xj/ � m�1j L. pn/ for all j D

1; : : : ; k. Since the sequence .L. pn//n2N converges, each sequence . pn.xj//n2Nis bounded. By passing to a subsequence if necessary, we can assume thataj WD limn pn.xj/ exists for all j. From pn 2 Pos.A;K/ we get aj � 0. Then

mi D L.xi/ D limn

L. pn/ D mi CkX

jD1;j¤i

mjaj: (18.27)

Since mj > 0 for all j by assumption, this implies that aj D limn pn.xj/ D 0 forall j D 1; : : : ; k; j ¤ i. Thus the sequence . pn/n2N has the desired properties.

(iii) From (ii), with pn D p; we get mi D L.xi/: Hence mi D L.xi/ by (i).(iv) Suppose that for x D xi the infimum in (18.23) is attained at p 2 A; that is,

L. p/ D L.xi/ and p.xi/ D 1. Thus we can set pn D p for n 2 N in the secondhalf of the proof of (ii). From (18.27) we conclude that aj D p.xj/ D 0 for allj ¤ i.

(v) Fix j 2 f1; : : : ; kg. Since mj D L.xj/, there is a sequence . pjn/n2N of elementsas stated in (i). To prove that k � dimA it suffices to show that the linearfunctionals lt1 ; : : : ; ltk on A are linearly independent. Assume that there arenumbers �1; : : : ; �k 2 R such that 0 D Pk

iD1 �ilxi. f / for all f 2 A. Settingf D pjn and passing to the limit n!1, we obtain

0 D limn

kXiD1

�ilxi. pjn/ DkX

iD1�i�

limn

pjn.xi/� D

kXiD1

�iıij D �j:

Thus, �j D 0 for all j, so the functionals lt1 ; : : : ; ltk are linearly independent. Inparticular, the points xj are pairwise distinct. Hence � is k-atomic. ut


The following definition is suggested by Proposition 18.31 (iii) and (iv).

Definition 18.32 We shall say that points x1; : : : ; xk 2 K have the positiveseparation property .PSP/A;K if the following holds:

There are elements q1; : : : ; qk 2 Pos.A;K/ such that qj.xi/ D ıij; i; j D 1; : : : ; k.

We summarize some of the preceding results in the following theorem.

Theorem 18.33 Let L be a truncated K-moment functional on A and let� D Pk

jD1 mjıxj be a k-atomic representing measure of L, where k 2 N andx1; : : : ; xk 2 K.

(i) If the points x1; : : : ; xk satisfy the positive separation property .PSP/A;K, thenmj D L.xj/ D L.xj/ for all j D 1; : : : ; k and the measure � is maximal mass.

(ii) Suppose that K is compact and the infimum (18.23) is attained at each pointx D xj; j D 1; : : : ; k: If � is maximal mass, then the points x1; : : : ; xk haveproperty .PSP/A;K.

(iii) Suppose thatK is compact and L is a relatively inner point of the cone L.A;K/.Then � is maximal mass if and only if the points x1; : : : ; xk obey the positiveseparation property .PSP/A;K.

Proof

(i) Fix j D 1; : : : ; k: Let qj be as in Definition 18.32. Setting p D qj in Proposition18.31(iii) we obtain mj D L.xj/ D L.xj/.

(ii) Since K is compact, L.xj/ D L.xj/ for all j by Proposition 18.27. Hencemj D L.xj/, because � is maximal mass. Therefore, Proposition 18.31(iv)implies that the points x1; : : : ; xk have property .PSP/A;K.

(iii) Since L is a relatively inner point of L.A;K/, it follows from Proposi-tion 18.28(iii) that the infimum in (18.23) is attained at each point x D xj.Now the assertion follows from (i) and (ii). ut

The next example gives a simple recipe for constructing maximal mass measures.

Example 18.34 (Squares of polynomials of degree one) Let x1; : : : ; xdC1 2 Rd.Assume that these points are not contained in a .d�1/-dimensional affine subspace,or equivalently, for any i 2 f1; : : : ; dC1g the d vectors xk � xi, k ¤ i, do not lie in a.d�1/-dimensional linear subspace of Rd.

Fix j 2 f1; : : : ; d C 1g and take an index i 2 f1; : : : ; d C 1g such that i ¤ j. Wechoose a vector aj 2 Rd, aj ¤ 0, which is orthogonal to the d � 1 vectors xk � xi,where k D 1; : : : ; dC 1, k ¤ j; i. Then aj � xk D aj � xi for k D 1; : : : ; dC 1, k ¤ j; i,where � is the scalar product of Rd. Further, aj is not orthogonal to xj � xi, sinceotherwise the vectors xk � xi; k ¤ i; would be contained in a .d � 1/-dimensionallinear subspace. Upon scaling aj we can assume that aj � .xj � xi/ D 1. Putting

qj.x/ WD .aj � x � aj � xi/2; (18.28)

we have qj.xi/ D ıij for i; j D 1; : : : ; d C 1 by construction.


Suppose that x1; : : : ; xdC1 2 K and RdŒ x �2 A. We define � D PdC1jD1 mjıxj

and L DPdC1jD1 mjlxj , where mj > 0. Since x1; : : : ; xdC1 obey the positive separation

property .PSP/A;K, the measure � 2ML;K is maximal mass by Theorem 18.33(i).ıExample 18.35 Let tj D .tj1; : : : ; tjd/, j D 1; : : : ; k; be k � 2 pairwise differentpoints of Rd. Since ti ¤ tj for i ¤ j, there is a number nij 2 f1; : : : ; dg such thatti;nij ¤ tj;nij . Then the Langrange interpolation polynomials

pj.x/ DkY

iD1;i¤j

xnij � ti;nijtj;nij � ti;nij

; j D 0; : : : ; k; (18.29)

satisfy pj 2 RdŒ x �k�1 and pj.ti/ D ıij; i; j D 1; : : : ; k.Suppose that K contains t1; : : : ; tk. Define � DPk

jD1 mjıtj and L DPkjD1 mjltj ;

where mj > 0 for j D 1; : : : ; k. Obviously, � 2ML;K:If RdŒ x �k�1 A, then � is an extreme point of ML;K by Proposition 18.10.Suppose that RdŒ x �2k�2 A. Then, since p2j 2 Pos.A;K/ and p2j .xi/ D ıij for

i; j D 1; : : : ; k, the atoms t1; : : : ; tk has the positive separation property .PSP/A;K.Therefore, � is maximal mass by Theorem 18.33(i). ı

18.5 Constructing Ordered Maximal Mass Measures

In this section, L is a truncatedK-moment functional on A such that L ¤ 0.Finding maximal mass representing measures is a difficult task even in the case

when K is compact. But there is a weaker notion of ordered maximal mass measuresfor which a simple general construction method exists.

Definition 18.36 A k-atomic measure � D PkiD1 miıti 2ML;K with all atoms ti

in K is called ordered maximal mass for L if

mj D Lj�1 .tj/ for j D 1; : : : ; k;

where Lj�1 is the functional on A defined by Lj�1 DPkiDj milti and L0 D L.

Obviously, each maximal mass measure is ordered maximal mass. The converseis not true; the measure � in Proposition 19.25 below is ordered maximal mass, butnot maximal mass.

From now on suppose thatK is compact. Then, from Proposition 18.27 we recallthat for each functional L0 2 L.A;K/ and t 2 K we have L0 .t/ D L0.t/ D cL0.t/and this number is the largest c 2 RC such that .L0 � clx/ 2 L.A;K/:

Since L ¤ 0 and there is an atomic measure in ML;K by Theorem 17.2, wehave L.t/ > 0 for some t 2 K. Set L0 WD L. We choose a point t1 2 K such thatL0 .t1/ > 0 and define L1 D L � L0 .t1/lx1 . Then L1 2 L.A;K/ and L1 .x1/ D 0.

18.5 Constructing Ordered Maximal Mass Measures 463

(Indeed, if L1 .x1/ > 0, there would exist a c > L0 .t1/ D cL0 .t1/ such that .L0 �clx/ 2 L.A;K/; a contradiction.) If L1 D 0, we set Lj D 0 for all j 2 N; j � 2, andwe are done. If L1 ¤ 0, we apply the preceding construction with L replaced by L1.Continuing this procedure, we obtain functionals Ln 2 L.A;K/ such that Ln D 0 orthere are points tj 2 K satisfying Lj�1 .tj/ > 0, Lj.tj/ D 0 for j D 1; : : : ; n, and

Ln D L �nX

jD1Lj�1 .tj/ltj : (18.30)

Let j � i � n. By construction, we have Li. f / � Lj. f / for f 2 Pos.A;K/. Hence0 � Li.tj/ � Lj.tj/ D Lj.tj/ D 0 by (18.22), so that Li.tj/ D Li.tj/ D 0.Therefore, if j < i, then j � i � 1 and hence Li�1 .tj/ D 0, but Li�1 .ti/ > 0, so thatti ¤ tj: In particular, this shows that the points t1; : : : ; tn are pairwise distinct.

Lemma 18.37 Lk D 0 for all k 2 N; k � jNj:Proof Assume to the contrary that Lk ¤ 0 for some k � jNj. Then LjNj ¤ 0: Setm WD jNj C 1. Since m > jNj D dimA, the point evaluations ltj , j D 1; : : : ;m; on Aare linearly dependent. Hence there are reals �1; : : : ; �m, not all zero, such that

mXjD1

�jltj . f / D 0; f 2 A: (18.31)

Let r be the smallest index for which �r ¤ 0: Since LjNj ¤ 0 and r � 1 � jNj, wehave Lr�1 ¤ 0 and Li�1 .ti/ > 0 for i D 1; : : : ;m by construction. From (18.30)it follows that Lr�1 D Lm C Pm

iDr Li�1 .ti/lti . Then, if �m is a finitely atomicrepresenting measure for Lm, so is �r�1 WD �m CPm

iDr Li�1 .ti/ıti for Lr�1 and�r�1.ftrg/ � Lr�1 .tr/. Hence �r�1.ftrg/ D Lr�1 .tr/ D Lr�1 .tr/ > 0: Thus,Proposition 18.31(ii) applies to �r�1, Lr�1, xi WD tr, so there exists a sequence. pn/n2N from Pos.A;K/ such that pn.tr/ D 1 and limn pn.t/ D 0 for all atoms t ¤ trof �r�1: Since Li�1 .ti/ > 0 for i D 1; : : : ;m; the points tj, r < j � m; are atoms of�r�1 and they are different from tr, as noted above. Setting f D pn in (18.31) andpassing to the limit we obtain

0 D limn!1

mXjD1

�j ltj. pn/ DmXjDr

�j�

limn!1 pn.tj/

� DmXjDr

�jıjr D �r:

This contradicts the choice of r and proves the assertion. utBy Lemma 18.37, the above construction terminates. Let k 2 N be the smallestnumber such that Lk�1 ¤ 0 and Lk D 0. By (18.30) we have for j D 1; : : : ; k,

L DkX

iD1Li�1 .ti/lti ; (18.32)

Lj�1 DkXiDj

Li�1 .ti/lti : (18.33)


Note that k � jNj by Lemma 18.37. From (18.32) it follows that the measure

� WDkX

jD1mjıtj ; where mj WD Lj�1 .tj/; j D 1; : : : ; k;

belongs to ML;K. Since the points t1; : : : ; tk are pairwise distinct, as shown above,and Lj�1 .tj/ > 0 for all j by construction, � is k-atomic. The functionalsLj�1 in (18.33) are precisely the linear functionals Lj�1 associated with � inDefinition 18.36. Therefore, � is an ordered maximal mass representing measurefor L according to Definition 18.36. Moreover, � can be chosen so that an arbitrarypoint t1 2 K satisfying L.t1/ > 0 is an atom of �. Then t1 2 VC.L;K/ byProposition 18.28(i). We summarize the preceding in the following theorem.

Theorem 18.38 Suppose that K is compact and L ¤ 0 is a truncated K-momentfunctional on A. Let t1 2 K be such that L.t1/ > 0: Then the above constructiongives a k-atomic measure � DPk

jD1mjıxj 2ML;K, k � jNj, such that all atoms of� are in K and � is ordered maximal mass.

Illustrative examples for this theorem are given in Sect. 19.4 and in Exercise19.11.

18.6 Evaluation Polynomials

Throughout this section, we assume that L is a fixed positive functional on A2, thatis, L is a real linear functional on A2 such that L. f 2/ � 0 for f 2 A:

From Proposition 17.17(v) and formula (17.30) we recall

NL D ff 2 A W L. f 2/ D 0 g; VL D fx 2 Rd W f .x/ D 0 for all f 2 NLg:

For f 2 A we denote by Qf D f CNL the equivalence class of f 2 A in DL D A=NL.Recall from Section 17.3 that there is a scalar product h�; �i on the vector space DL

defined by L. fg/ D hQf ; Qgi; f ; g 2 A:Let t 2 VL: Since f .t/ D 0 for f 2 NL, there is a well-defined linear functional

Ft on DL D A=NL given by Ft. Qf / WD f .t/; f 2 A: Each linear functional on thefinite-dimensional real Hilbert space .DL; h�; �i/ is continuous, so by Riesz’ theoremthere is a unique element eEt 2 DL, Et 2 A, such that

f .t/ � Ft. Qf / D hQf ; eEti D L. f Et/; f 2 A: (18.34)

Definition 18.39 Et 2 A is called an evaluation polynomial at the point t 2 VL:

Note that the polynomial Et is not uniquely determined by t. In fact, it is onlydetermined up to a summand from the vector space NL, but the values of Et.�/ onthe set VL do not depend on the particular choice of the element of the equivalenceclass eEt. In what follows we will fix a choice of the polynomial Et.

18.6 Evaluation Polynomials 465

Let fQfi W i D 1; : : : ; kg be an orthonormal basis of the Hilbert space DL.Using (18.34) we develop eEt with respect to this basis and obtain

eEt DkX

iD1heEt; Qfii Qfi D

kXiD1

fi.t/Qfi: (18.35)

As noted above, Qf .x/ D f .x/ for x 2 VL and f 2 A. Therefore,

Et.x/ DkX

iD1fi.t/fi.x/ for t; x 2 VL: (18.36)

Since the fi are polynomials, it follows from (18.36) that Et.x/ is jointly continuousin .t; x/ for t; x 2 VL: Using (18.34) we derive

Et.t/ D L.EtEt/ D h QEt; QEti D k QEtk2 for t 2 VL; (18.37)

where k � k denotes the norm of the Hilbert space DL. In particular, eEt ¤ 0 in DL

and hence Et.t/ ¤ 0, since he; eEti D e.t/ > 0 by (18.34) and (18.1).For t 2 Rd we now define another quantity �L.t/ by a variational problem:

�L.t/ D inf fL.p2/ W p 2 A; p.t/ D 1g D inf

�L. p2/

p.x/2W p 2 A

�; (18.38)

where we set c0WD C1 for c � 0. The following result is similar to Proposi-

tion 9.12.

Proposition 18.40 For t 2 VL the infimum in (18.38) is a minimum and equal tokeEtk�2. Further, keEtk�2eEt is the unique element Qp 2 DL such that the infimum in(18.38) is attained at p 2 A.

Proof The proof is almost verbatim the same as the proof of Proposition 9.12.Let p 2 A: We develop Qp with respect to the orthonormal basis fef1; : : : ;efkg:

Qp DX

iciefi; where ci D hQp;efii; and L. p2/ D kQpk2 D

Xic2i :

Hence p.t/ DPi cifi.t/. Therefore, the problem in (18.38) is to minimizeP

i c2i for

.c1; : : : ; ck/T 2 Rk under the constraintP

i cifi.t/ D 1. This problem is solved byLemma 9.13, now applied to Rk with g D .c1; : : : ; ck/T and f D . f1.t/; : : : ; fk.t//T .Since fef1; : : : ;efkg is an orthonormal basis, (18.35) yields keEtk2 D P

i fi.t/2. Then,

by Lemma 9.13, the infimum is attained at p if and only if ci D fi.t/keEtk�2, that is,

Qp DX

ifi.t/keEtk�2efi D keEtk�2eEt;

and the minimum is�P

i fi.t/2��1 D keEtk�2. ut


Note that if L is a truncated K-moment functional, then we have

�L.t/ � L.t/ � L.t/ for t 2 VL \K: (18.39)

Indeed, the first inequality follows at once by comparing the corresponding defini-tions (18.22) and (18.38), while the second inquality holds by (18.25).

Definition 18.41 A family fExj W j D 1; : : : ; kg of evaluation polynomials Exj , xj 2VL, is called orthogonal if hfExi ;fExji D 0 for i; j D 1; : : : k; i ¤ j.

As noted above, fExi ¤ 0. Hence a set fExj W j D 1; : : : ; kg of evaluationpolynomials is orthogonal if and only if there are numbers ci > 0, i D 1; : : : ; k;such that

hfExi ;fExji D c�1

i ıij; i; j D 1; : : : ; k: (18.40)

Here the inverse c�1i is taken in order to obtain formula (18.41) below. Note that

combining (18.34) and (18.40) yields Exi.xj/ D c�1i ıij.

The remaining results of this section deal with truncated moment functionals Lthat have .rankL/-atomic representing measures. The next theorem characterizessuch functionals in terms of evaluation polynomials.

Theorem 18.42 Let L be a positive functional on A2. Set k WD rankL. There isa one-to-one correspondence between orthogonal families fExj W j D 1; : : : ; kg ofevaluation polynomials Exj , xj 2 VL; and k-atomic representing measures � of L. It

is given by � DPkjD1 cjıxj , where cj are the numbers from (18.40). In this case,

�L.xj/ D cj ; j D 1; : : : ; k: (18.41)

Proof Let fExj W j D 1; : : : ; kg be an orthogonal family of evaluation polynomials.

Put � D PkjD1 cjıxj . Since k D rankL D dim DL by Proposition 17.17(iii), it

follows from (18.40) that fc1=2jeExj W j D 1; : : : ; kg is an orthonormal basis of the

Hilbert space DL. Therefore, using Parseval’s identity and (18.35) we derive

L. pq/ D h Qp; Qqi DX

jh Qp; c1=2j

eExj ihQq; c1=2jeExji

DX

jcjh Qp;eExji hQq;eExji D

Xjcjp.xj/q.xj/ D

Z. pq/.x/ d�.x/

for p; q 2 A. Since A2 is the span of products pq with p; q 2 A, the preceding impliesthat L. f / D R f d� for all f 2 A2, that is, � 2ML.

Conversely, suppose that � D PkiD1miıxi is a k-atomic measure of ML. By

Proposition 17.17(vi), each atom xj is in VL. Let J denote the embedding of A intoL2.Rd; �/ and h�; �i� the scalar product of L2.Rd; �/. Then

h Qp; Qqi D L. pq/ DZ

pq d� D hJ. p/; J.q/i� ; p; q 2 A: (18.42)

18.6 Evaluation Polynomials 467

Hence Qp 7! J. p/ is a well-defined isometric linear map of DL into L2.Rd; �/. Since� is k-atomic and hence dimL2.Rd; �/ D k D dim DL, J is bijective. Hence forany j D 1; : : : ; k there is a pj 2 A such that J. pj/ is the characteristic function of thepoint xj, that is, J. pj/.xi/ D pj.xi/ D ıji for i; j D 1; : : : ; k: By (18.42),

hQf ;m�1j epji D m�1

j L. fpj/ D m�1j

Zfpj d� D m�1

j

Ximif .xi/pj.xi/ D f .xj/

for f 2 A, that is, Exj WD m�1j pj is an evaluation polynomial at xj. Further,

hm�1i epi;m�1

j epjiDm�1i m�1

j

Zpipj d� D m�1

i m�1j

Xnmnpi.xn/pj.xn/ D m�1

i ıij:

This shows that fExj W j D 1; : : : ; kg is an orthogonal family of evaluationpolynomials for L with constants cj D mj in (18.40). The corresponding measure isP

j mjıxj D �. This proves the asserted one-to-one correspondence.We verify (18.41). Take p 2 A such that p.xj/ D 1. Since k D dim DL, the set

feExj W j D 1; : : : ; kg is an orthogonal basis of DL; so Qp D PkiD1 ˛ieExi with ˛i 2 R.

By (18.40), we have keExi k2 D Exi.xi/ D c�1i for each i. Using this formula we

derive 1 D p.xj/ DPkiD1 ˛iExi.xj/ D ˛jc�1

j , so that ˛j D cj for all j. Therefore,

L. p2/ D k Qp k2 DX

i˛2i keExi k2 D

Xi˛2i c

�1i :

Hence the infimum in Eq. (18.38) is obviously obtained when we set ˛i D 0 for alli ¤ j. Thus, �L.xj/ D ˛2j c�1

j D cj. utThe next theorem deals with the “nice” but rare case when Pos.A2;K/ DPA2; seee.g. Exercise 18.9 for a very simple example.

Theorem 18.43 Suppose that the set K is compact and

Pos.A2;K/ DX

A2 WD ˚ Xif 2i W fi 2 A

�:

(i) If L is a positive linear functional on A2, then L has a .rankL/-atomic maximalmass representing measure � with all atoms contained in VL \K. Moreover,

L.x/ D �L.x/ for x 2 VL \K: (18.43)

(ii) Assume that f dK D 0 for f 2 A implies f D 0. Then C.A2;K/ D jNj:Proof

(i) First we note that L is a truncated K-moment functional by Theorem 17.10,since Pos.A2;K/ DPA2 and K is compact.


We verify (18.43). Define L1 WD L � �L.x/lx: By the second equalityin (18.38), we have L1. f 2/ D L. f 2/ � �L.x/f .x/2 � 0 for f 2 A; thatis, L1 is a positive functional on A2. Thus, again by Theorem 17.10, L1 is atruncated K-moment functional. If �1 2ML1;K; then � WD �1 C �L.x/ıx is arepresenting measure of L D L1 C �L.x/lx: Hence L.x/ � �.fxg/ � �L.x/.Since L.x/ � �L.x/ by (18.39), (18.43) is proved.

Next we prove the following assertion: If �L.x/ > 0, then

rankL D 1C rankL1: (18.44)

Indeed, since L. f 2/ � L1. f 2/ � 0 for f 2 A, we have Hn.L/ � Hn.L1/�0 andhence rankL � rankL1. By Proposition 18.40 there is a uniqueep 2 DL suchthat the infimum in (18.38) is attained at p, that is, p.x/ D 1 and �L.x/ D L. p2/.Then L1. p2/ D 0 and L. p2/ ¤ 0. From the uniqueness of Qp and (17.29) weconclude that NL is a subspace of NL1 of codimension 1. Hence (18.44) followsfrom Proposition 17.17.

To prove the remaining assertions on L we shall proceed by induction.First assume that �L.x/ D 0 for all x 2 VL \ K. Then L.x/ D 0 on VL \ K

by (18.43). Since each� 2ML;K is supported on VL\K, Theorem 17.2 impliesthat L D 0, so the assertions on L hold trivially.

Now we assume that �L.x1/ > 0 for some x1 2 VL \ K. Then, asshown above, L1 WD L � �L.x1/lx1 is a truncated K-moment functional satis-fying (18.44). We replace L by L1 and proceed by induction. After k WD rankLsteps we obtain a positive functional Lk such that rankLk D 0. Then Lk D 0 and

L D Lk CkX

jD1�Lj�1 .xj/lxj D

kXjD1

�Lj�1 .xj/lxj ; where L0 WD L:

Hence � WDPkjD1 �Lj�1 .xj/ıxj is a representing measure of L such that

jsupp�j � k D rankL D rankH.L/:

By (17.38), rankH.L/ � jsupp�j: Thus rankL D jsupp�j; so � is.rankL/-atomic. Therefore Theorem 18.42 applies and formula (18.41) yields�Lj�1 .xj/ D �L.xj/ for j D 1; : : : ; k. Since �L.xj/ D L.xj/ by (18.43), we have�Lj�1 .xj/ D L.xj/, that is, � has maximal mass at each atom xj. Thus, � ismaximal mass.

(ii) By (i), each L 2 L.A2;K/ has a .rankL/-atomic representing measure. HenceC.A2;K/ � jNj; since rankL � jNj. From the assumption of (ii) it followsthat there are points x1; : : : ; xjNj 2 K such that the functionals lx1 ; : : : ; lxjNj

are

linearly independent on A. Define L D PjNjjD1 lxj and � D PjNj

jD1 ıxj : ThenL 2 L.A2;K/ and � 2ML;K: From Proposition 17.21, rankL D rankH.L/ DjNj. By (17.38), L has no representing measure with fewer atoms. Hence jNj �C.A2;K/. Thus, C.A2;K/ D jNj: ut

18.7 Exercises 469

Remark 18.44 In condition (18.1) we assumed that A contains an element e suchthat e.x/ � 1 on K. If L is a truncated K-moment functional on A2, all results ofthis section hold if e 2 A2 rather than e 2 A: Indeed, the proof of Theorem 18.43remains valid, since Theorem 17.10 was applied to A2. Since e 2 A2, for each x 2 Kthere is an fx 2 A such that fx.x/ ¤ 0; so hfx; eExi D fx.x/ ¤ 0, and hence eEx ¤ 0: ı

There is a similar result about positive functionals of “small rank”. It is based onthe following theorem due to G. Blekherman [Bl2, Theorem 2.3].

Theorem 18.45 If L is a positive linear functional on RdŒ x �2n such that

rank L � 3n � 3 if n � 3; rank L � 6 if n D 2; (18.45)

then L is a truncated moment functional.

The proof in [Bl2] is based on tools from algebraic geometry (Cayley–Bacharachduality, see e.g. [EGH]) which are beyond the scope of this book. Taking Theo-rem 18.45 for granted we obtain the following result.

Theorem 18.46 Let d; n 2 N and let L be a positive linear functional on RdŒ x �2n;that is, L. f 2/ � 0 for f 2 RdŒ x �n. Suppose that condition (18.45) is satisfied. ThenL is a truncated moment functional with .rankL/-atomic representing measure andthe equality (18.43) holds with K D Rd.

The proof of Theorem 18.46 follows the same lines as the proof of Theo-rem 18.43 with Theorem 17.10 replaced by Theorem 18.45. We omit the details.

18.7 Exercises

1. Suppose that K is compact and let f 2 A. Prove the following:

a. f is an interior point of Pos.A;K/ if and only if f .x/ > 0 for all x 2 K:b. f is a boundary point of Pos.A;K/ if and only if f .x0/ D 0 for some x0 2 K:c. The assertion of a. is no longer valid in general if K is not compact.

Hint: f .x/ D 1C x2;K D R;A D RŒx�n with n � 3:2. Let A D RdŒ x �2n and let K be a closed subset of Rd. Discuss whether or not

(depending on K) the following statements are true:

a. Each boundary point of S.A;K/ is determinate.b. Each interior point of S.A;K/ is indeterminate.

3. ( NC.Ls;K/ and VC.Ls;K/ for one-dimensional bounded intervals)Let d D 1;K D Œ0; 1�; m 2 N; and s 2 SmC1; s ¤ 0. Suppose that ind.s/ �

m, see Definitions 10.4 and 10.6. Use Theorem 10.7 to determine NC.Ls;K/ andVC.Ls;K/.


4. Let d D 1;K D Œ0; 1�;A D RŒx�2; and x1 D 0; x2 D 12; x3 D 1. Define L DP3

jD1 lxj . Show that L.0/ D 65.

5. Let d D 1;K D Œ0; 1�;A D RŒx�n. Consider a k-atomic measure � DPkjD1mjıxj

and L DPkjD1mjlxj :

a. Decide when mj D L.xj/ for one j.b. Decide when � is maximal mass.c. Give an example such that m1 D L.x1/, but m2 ¤ L.x2/.d. Discuss these problems in the case K D R.

6. Let d D 2, x1 D .0; 0/; x2 D .1; 0/; x3 D .0; 1/; x4 D .1; 1/. Set � WDP4jD1mjıxj and L WD P4

jD1mjıxj , where mj > 0; j D 1; 2; 3; 4: Suppose thatx1; x2; x3; x4 2 K.

a. Suppose that R2Œ x �2 A. Show that � is an extreme point of ML;K:b. Let K be a rectangle and A D R2Œ x �2. Find a 3-atomic measure � 2ML;K:c. Suppose that R2Œ x �4 A. Show that � is determinate.

7. Suppose that K is compact. Show that C.A;K/ � 1C [email protected];K/C.s/:Hint: Use Proposition 1.43. Write s 2 S.A;K/ as s D s0 C cLs.x/s.x/, s0 [email protected];K/:

8. Extend the construction of maximal mass measures in Example 18.34 by takingproducts of squares of linear polynomials in (18.28).

9. Let d � 2, N D f.1; 0; : : : ; 0/; : : : ; .0; : : : ; 0; 1/g: Let K be a compact subset ofRd such that Sd�1 K. Then A2 are the homogeneous polynomials in RdŒ x � ofdegree 2. Show that Pos.A2;K/ DP A2 and C.A2;K/ D d.

18.8 Notes

The real algebraic set VC.L/ first appeared in the Thesis of J. Matzke [Mt], whilethe core variety V.L/ was invented by L. Fialkow [F2]. Theorem 18.22 is due to G.Blekherman and L. Fialkow (personal communication from L. Fialkow, June 2016).Theorems 18.21 and 18.23 were proved in [DSm1]. Most of the results on maximalmasses in Sects. 18.4 and 18.5 are taken from [Sm10]. A number of results of thischapter (for instance, Theorems 18.42 and 18.43) are new.

Chapter 19The Truncated Moment Problemfor Homogeneous Polynomials

Let Hd;m denote the real homogeneous polynomials in d variables of degreem. In thefinal chapter of the book we deal with the truncated moment problem for Hd;2n onRd, on the unit sphere Sd�1, and on Sd�1C . Since Sd�1C is a realization of the projectivespace Pd�1.R/, the latter is in fact the truncated projective moment problem. Theexistence problem in this case was considered in Sect. 17.2.

In Sects. 19.1 and 19.2, we treat the apolar scalar product on the vector spaceHd;m and its relations to the actions of differential operators. These are powerfultechnical tools for the study of homogeneous polynomials. They are appliedin Sect. 19.3 to the truncated projective moment problem and to Carathéodorynumbers. In Sect. 19.4, we develop some properties of the Robinson polynomialand use them to construct interesting examples on the truncated moment problem.In Sect. 19.5, some results on zeros of positive polynomials (Theorem 19.32) arederived. Section19.6 contains some applications to the truncated moment problemon R2.

In this chapter, N is the set

Nd;m WD f˛ 2 Nd0 W ˛1 C � � � C ˛d D mg; where m 2 N;

and the corresponding span A D Lin fx˛ W ˛ 2 Nd;m g is the vector space Hd;m. Theelements of Hd;m are also called d-forms of degree m, or briefly, forms.

19.1 The Apolar Scalar Product

First we note that the vector space Hd;m has the dimension

dimHd;m D jNd;mj D mC d � 1d � 1

!: (19.1)


471

472 19 The Truncated Moment Problem for Homogeneous Polynomials

Our first aim is to define the apolar scalar product Œ�; �� on Hd;m. We abbreviate

m

˛

!D mŠ

˛1Š : : : ˛dŠfor ˛ D .˛1; : : : ; ˛d/ 2 Nd;m:

Let us consider homogeneous polynomials

p.x/ DX˛2Nd;m

m

˛

!a˛x

˛ and q.x/ DX˛2Nd;m

m

˛

!b˛x

˛ (19.2)

of Hd;m and define

Œ p; q� WDX˛2Nd;m

m

˛

!a˛b˛: (19.3)

(The reason for including the multinomial coefficients�m˛

�in (19.2) and (19.3) will

be seen below when we derive a number of nice formulas.)

Definition 19.1 Œ�; �� is called the apolar scalar product on Hd;m.

It is obvious that Œ�; �� is a scalar product on Hd;m. Thus, .Hd;m; Œ�; ��/ is a finite-dimensional real Hilbert space. Note that the coefficient a˛ of p 2 Hd;m in (19.2)can be recovered from the apolar scalar product by

Œ p; x˛� D a˛; ˛ 2 Nd;m:

As usual, a � b D a1b1 C � � � C adbd denotes the standard scalar product of Rd.Let y D .y1; : : : ; yd/ 2 Rd. We denote by .y�/m the element of Hd;m defined by

.y �/m.x/ WD .y � x/m �� dX

jD1yjxj

�m

DX˛2Nd;m

m

˛

!y˛x˛; (19.4)

where the last equality holds by the multimonomial theorem. The map y 7! .y � x/2n,called the 2n-th Veronese embedding, plays an important role in algebraic geometry.

Now suppose that f .x/ D PkjD1 cj.yj � x/m 2 Hd;m, where yj 2 Rd and cj 2 R,

and let p.x/ 2 Hd;m be as in (19.2). From (19.4) it follows that

Œ p; f � DkX

jD1

m

˛

!a˛cjy

˛j D

kXjD1

cjp.yj/ DkX

jD1cjlyj. p/; p 2 Hd;m: (19.5)

Thus, the scalar product with f is just the corresponding linear combination of pointevaluations lyj at yj. This gives the link to the truncated moment problem. Further,

19.1 The Apolar Scalar Product 473

Eq. (19.5) shows the reproducing kernel property of the apolar scalar product:

Œ p; .y � x/m� D ly. p/ D p.y/; y 2 Rd; p 2 Hd;m: (19.6)

That is, the point evaluation at y is the linear functional given by .y�/m.In particular, setting p D .a�/m and y D b in (19.6), we obtain

Œ.a �/m; .b �/m� D .a � b/m; a; b 2 Rd; (19.7)

where a � b means the usual scalar product in Rd. Therefore, if a?b in Rd, then.a �/m?.b �/m in the Hilbert space .Hd;m; Œ�; ��/. Further, if the set fy1; : : : ; yrg isorthonormal in Rd, so is f.y1 �/m; : : : ; .yr �/mg in .Hd;m; Œ�; ��/.

Next we derive some useful facts on bases and spanning sets of Hd;m.

Lemma 19.2 For each open subset U ofRd, the polynomials .y �/m.x/; y 2 U; spanthe vector spaceHd;m:

Proof Let p 2 Hd;m: Since .Hd;m; Œ�; ��/ is a Hilbert space, it suffices to show thatŒ p; .y � x/m� D 0 for all y 2 U implies p D 0. By (19.6), Œ p; .y � x/m� D p.y/ D 0

on the open set U. Therefore, p D 0: utThe following is a classical result of O. Biermann (1903).

Proposition 19.3 The set f.˛�/m.x/ W ˛ 2 Nd;mg is a basis of the vector spaceHd;m:

Proof For ˇ 2 Nd;m we define

fˇ.x/ DdY

jD1

ˇj�1YiD0

.mxj � i.x1 C � � � C xd//: (19.8)

Since fˇ is a product of jˇj D m homogeneous linear polynomials, fˇ 2 Hd;m:

Let ˛ 2 Nd;m: First suppose that ˛ ¤ ˇ. There is an index j such that ˛j < ˇj.Then there is a factor with indices j; i D ˛j in (19.8). Evaluated at the point x D ˛

this factor is m˛j � ˛j.˛1 C � � � C ˛d/ D 0. Hence fˇ.˛/ D 0. If ˇ D ˛, then

fˇ.ˇ/ DdY

jD1

ˇj�1YiD0

.mˇj � i m/ D mmdY

jD1ˇjŠ ¤ 0:

Thus, Œ fˇ; .˛ � x/m� D fˇ.˛/ D 0 for ˛ ¤ ˇ and Œ fˇ; .ˇ � x/m� D fˇ.ˇ/ ¤ 0.This in turn implies that the polynomials .˛�/m.x/, where ˛ 2 Nd;m; are linearlyindependent. Since the vector space Hd;m D Lin fx˛ W ˛ 2 Nd;mg has dimensionjNd;mj, they form a vector space basis of Hd;m. utDefinition 19.4 A subset fy1; : : : ; yrg of Rd is called a basic set of nodes if thepolynomials f.y1 �/m; : : : ; .yr �/mg form a basis of the vector space Hd;m.


Obviously, if fy1; : : : ; yrg is a basic set of nodes, then r D jNd;mj D dimHs;m.Proposition 19.3 says that the set Nd;m is a basic set of nodes.

Remark 19.5 By Lemma 19.2, each polynomial f 2 Hd;m can be written in the formf .x/ D Pk

jD1 cj.yj � x/m, where cj 2 R and yj 2 Rd. The smallest number k amongall such representations of f is called the real Waring rank of f . Note that here thenumbers cj are real, while for the width w. f / of f 2 Qd;2n (see Definition 19.17below) the numbers cj are positive. ı

Let f fi W i 2 Ig and fgi W i 2 Ig be subsets of Hd;m, where I is an index set suchthat jIj D dimHd;m D

�mCd�1d�1

�: We say that the sets are dual bases of Hd;m if

Œ fi; gj� D ıi;j for i; j 2 I: (19.9)

It is well-known from linear algebra and it is easy to verify that in this case the setsf fi W i 2 Ig and fgi W i 2 Ig are indeed vector space bases of Hd;m and

f DX

i2I Œ fi; f � gi DX

i2I Œgi; f � fi for f 2 Hd;m: (19.10)

Proposition 19.6 Two subsets f fi W i 2 Ig and fgi W i 2 Ig, jIj D �mCd�1d�1

�, of Hd;m

form dual bases ofHd;m if and only if the following Marsden identity holds:

.y � x/m DX

i2I fi.y/gi.x/; x; y 2 Rd: (19.11)

Proof Suppose that these sets are dual bases. Setting f D .y �/m in (19.10) and usingthat Œ fi; .y �/m� D fi.y/ by (19.6) we obtain

.y �/m DXi

fi.y/gi: (19.12)

Evaluating both sides of this equation at x yields (19.11).Conversely, assume that (19.11) holds. Then (19.12) holds for y 2 Rd. Since

Hd;m is spanned by the functions .y�/m, y 2 Rd; (19.12) implies that the functionsgi; i 2 I; span the whole vector space Hd;m: Therefore, since jIj D �mCd�1

d�1� D

dimHd;m, it follows that fgi W i 2 Ig is a basis of Hd;m. Now we take y as a variableand apply the apolar scalar product with gj.y/ in (19.11). Then we obtain

gj.x/ D Œgj.y/; .y � x/m� DX

i2I Œ fi; gj�gi.x/; x 2 Rd:

Since fgi W i 2 Ig is a basis, we conclude that Œ fi; gj� D ıij for i; j 2 I. This meansthat the corresponding sets are dual bases. ut

19.2 The Apolar Scalar Product and Differential Operators 475

19.2 The Apolar Scalar Product and Differential Operators

The apolar scalar product is a valuable tool for the study of forms and truncatedmoment problems. One reason for this is its nice interplay with differentialoperators.

Let i 2 f1; : : : ; dg and ˛ D .˛1; : : : ; ˛d/ 2 Nd0: Then we abbreviate @i D @

@xiand

@˛ D .@1/˛1 � � � .@d/˛d . Further, we define

p.@/ WDX

˛a˛@

˛ for p.x/ DX

˛a˛x

˛ 2 RdŒ x �

and consider p.@/ as a differential operator acting on polynomials. Since theoperators @

@xiand @

@xjcommute on polynomials, we have

.pg/.@/f D p.@/q.@/f D q.@/p.@/f for p; q; f 2 RdŒ x �: (19.13)

Lemma 19.7 Let y1; : : : ; yk 2 Rd, c1; : : : ; ck 2 R, and define

f .x/ DkX

jD1cj.yj � x/m 2 Hd;m:

If p 2 Hn;d and n � m, then

. p.@/f /.x/ D m.m � 1/ : : : .mC 1 � n/kX

jD1cjp.yj/.yj � x/m�n: (19.14)

Let y 2 Rd and p 2 Hd;m. Then

p.@/.y � x/m D mŠ p.y/: (19.15)

In particular, if p.y/ D 0, then p.@/.y � x/m D 0:Proof For ˛ 2 Nd;n we derive

@˛.y � x/m D�@

@x1

�˛1� � ��@

@xd

�˛d.y � x/m

D m.m � 1/ : : : .mC 1 � n/ y˛11 � � � y˛dd .y � x/m�n

D m.m � 1/ : : : .mC 1 � n/ y˛.y � x/m�n:

By linearity this implies (19.14). Formula (19.15) follows from (19.14) by settingf .x/ D .y � x/m. utLemma 19.8 If p; q 2 Hd;m; then Œ p; q� D 1

mŠp.@/q D 1mŠq.@/p:


Proof By the symmetry of the scalar product it suffices to prove the first formula.By linearity we can restrict ourselves to p D �m

˛

�x˛ and q D xˇ , where ˛; ˇ 2 Nd;m.

First suppose that ˛ ¤ ˇ. Then Œx˛; xˇ� D 0 by (19.3). Since ˛; ˇ 2 Nd;m and

˛ ¤ ˇ, there is an index j 2 f1; : : : ; dg such that ˛j > ˇj. Then . @@xj/˛j x

ˇjj D 0 and

hence @˛xˇ D 0. That is, Œ�m˛

�x˛; xˇ� D 1

mŠ

�m˛

�@˛xˇ D 0 for ˛ ¤ ˇ.

Now assume that ˛ D ˇ. Then we have ˛j D ˇj and hence . @@xj/˛j x

ˇjj D ˛jŠ for

all j D 1; : : : ; d. This yields @˛xˇ D ˛1Š : : : ˛dŠ and

1

mŠ

m

˛

!@˛xˇ D 1

mŠ

mŠ

˛1Š : : : ˛dŠ˛1Š : : : ˛dŠ D 1 D

� m˛

!x˛; xˇ

�:

This completes the proof. utCombining formula (19.15) with Lemma 19.8 we obtain the following

Corollary 19.9 If f .x/ DPkjD1 cj.yj � x/m 2 Hd;m with yj 2 Rd, cj 2 R, then

1

mŠf .@/p D Œ p; f � D

kXjD1

cjp.yj/ DkX

jD1cjlyj. p/ for p 2 Hd;m: (19.16)

Formula (19.16) expresses the apolar scalar product in terms of differentialoperators. In Theorems 19.14 and 19.16 below we apply this to the momentproblem.

Lemma 19.10 Let n 2 N; n < m. If p 2 Hd;m, q 2 Hd;n and f 2 Hd;m�n, then

mŠŒ p; fq�m D .m � n/ŠŒ f ; q.@/p�m�n;

where Œ�; ��m and Œ�; ��m�n are the apolar scalar products of Hd;m and Hd;m�n,respectively.

Proof Using Lemma 19.8 and Eq. (19.13) we derive

mŠŒ p; fq�m D .fq/.@/p D f .@/.q.@/p/ D .m � n/ŠŒ f ; q.@/p�m�n: ut

Remark 19.11 Let n;m 2 N; n � m. From formula (19.14) it follows that

Œ p; f �n;m WD .m.m � 1/ : : : .mC 1 � n//�1p.@/f ; p 2 Hd;n; f 2 Hd;m; (19.17)

defines a bilinear mapping, called the apolar pairing, of Hd;n �Hd;m to Hd;m�n. Inthe special case m D n the right-hand side of (19.17) becomes the scalar mŠp.@/fand we obtain the apolar scalar product Œ�; �� on Hd;m. ı

We close this section by deriving the following version of Sylvester’s apolaritylemma. It illustrates nicely the usefulness of the apolar scalar product, but will notbe needed later for the study of moment problems.

19.3 The Apolar Scalar Product and the Truncated Moment Problem 477

Proposition 19.12 Let k;m 2 N; k � m; and .aj; bj/ 2 R2, j D 1; : : : ; k: Supposethat ajbi ¤ aibj for i ¤ j. Set q D .b1x1 � a1x2/ : : : .bkx1 � akx2/: Let f 2 H2;m:

Then we have q.@/f D 0 if and only if f is of the form

f DkX

jD1cj.ajx1 C bjx2/

m with c1; : : : ; cm 2 R: (19.18)

Proof Let E be the subspace of H2;m spanned by hj WD .ajx1C bjx2/m, j D 1; : : : ; k.We describe the orthogonal compliment E? of E with respect to the apolar scalar

product. Let g 2 H2;m. Then g 2 E? if and only if Œhj; g� D g.aj; bj/ D 0 forj D 1; : : : ; k. Thus, E? is the set of g 2 H2;m vanishing at all .aj; bj/. Therefore,since g is a homogeneous polynomial in 2 variables, each polynomial bjx1 � ajx2divides g. The assumption ajbi ¤ aibj for i ¤ j implies that bix1�aix2 and bjx1�ajx2are relatively prime for i ¤ j. Hence the product q of all factors bjx1 � ajx2 dividesg. That is, g D hq for some h 2 H2;m�k. Conversely, each g of the form g D hqvanishes at all points .aj; bj/, so it is in E?. Summarizing, we have shown that

E? D fhq W h 2 H2;m�k g: (19.19)

First let q.@/f D 0. Then mŠŒg; f � D g.@/f D h.@/q.@/f D 0 for g D hq 2 E?by (19.19), so that f 2 E?? D E. This means that f is of the form (19.18).

Conversely, assume that f has a representation (19.18). Then f 2 E. By (19.19),mŠŒhq; f � D .hq/.@/f D h.@/q.@/f D 0 for all h 2 H2;m�k. If k D m, thenq.@/f D 0 and we are finished. Now suppose k < m: Note that q.@/f 2 H2;m�k:

If Œ�; ��0 denotes the apolar scalar product of H2;m�k, we have .m � k/ŠŒh; q.@/f �0 Dh.@/q.@/f D 0 for all h 2 H2;m�k, that is, q.@/f 2 .H2;m�k/

? D f0g. Thus,q.@/f D 0. ut

19.3 The Apolar Scalar Product and the Truncated MomentProblem

Let us begin by defining three cones in the vector space Hd;m resp. Hd;2n:

Qd;m WD˚f 2 Hd;m W f D

kXjD1.yj�/m; where y1; : : : ; yk 2 Rd; k 2 N

�; (19.20)

X2

d;2nWD ˚ f 2 Hd;2n W f D

kXjD1

f 2j ; where f1; : : : ; fk 2 Hd;n; k 2 N�; (19.21)

Pd;2n WD Pos.Hd;2n;Rd/ � ˚f 2 Hd;2n W f .x/ � 0 for x 2 Rd

�: (19.22)


Since �.y � x/m D . mp�y � x/m for � � 0, Qd;m is a cone in Hd;m. Obviously, Pd;2n andP2

d;2n are cones in Hd;2n. Clearly,

Qd;2n X2

d;2n Pd;2n: (19.23)

In general, we have Qd;2n ¤ Pd;2n. Exercise 19.6 contains an example of apolynomial p 2 P2

2;4 such that p.x/ > 0 for all x 2 R2nf0g, but p … Q2;4: At

the end of this section we discuss briefly when the equalityP2

d;2n D Pd;2n holds.Recall from Lemma 19.2 that the polynomials .y�/m 2 Qd;m; y 2 Rd, span Hd;m.

Thus Qd;m, henceP2

d;2n and Pd;2n by (19.23), span Hd;m resp. Hd;2n. Therefore, allthree cones have nonempty interiors in Hd;m resp. Hd;2n by Proposition A.33(i).

Proposition 19.13 Qd;2n and Pd;2n are closed in the norm topology of Hd;2n andthey are dual cones to each other with respect to the apolar scalar product, that is,

Qd;2n D f f 2 Hd;2n W Œ p; f � � 0 for p 2 Pd;2n g; (19.24)

Pd;2n D f p 2 Hd;2n W Œ p; f � � 0 for f 2 Qd;2n g: (19.25)

Proof Obviously, Pd;2n is closed. We prove that Qd;2n is closed. Set r WD dimHd;2n.Let . fk/k2N be a sequence from Qd;2n converging to f 2 Hd;2n. From Carathéodory’stheorem (Proposition A.35) and the definition of Qd;2n it follows that each fk is ofthe form

fk.x/ DrX

jD1.y.k/j � x/2n; where y.k/j D .ykj1; : : : ; ; ykjd/T 2 Rd: (19.26)

Fix i 2 f1; : : : ; dg. The coefficient of x2ni in fk isPr

jD1.ykji/2n. Clearly, as k ! 1,these numbers converge to the coefficient of x2ni in f . In particular, these sequencesare bounded, so there exists a C > 0 such that jykjij � 1CPr

lD1.ykli/2n � C for allk; j; i. Hence we can choose a subsequence .km/m2N such that yji WD limm!1 ykmjiexists. Setting yj D .yj1; : : : ; yjd/T and passing to the limit k!1 in (19.26) we getf .x/ DPr

jD1.yj�/2n, so that f 2 Qd;2n. This proves that Qd;2n is closed.We verify (19.24) and (19.25). We identify the dual of Hd;2n with the functionals

Œ�; f � and Œ p; ��, respectively; then the sets on the right-hand sides of (19.24)and (19.25) are the dual cones .Pd;2n/

^ and .Qd;2n/^, respectively. Let f DPr

jD1.yj�/2n and p 2 Pd;2n: Then, by (19.5), Œ p; f � D PrjD1 p.yj/ � 0. Therefore,

Qd;2n .Pd;2n/^ and Pd;2n .Qd;2n/

^; so that .Qd;2n/^ � .Pd;2n/

^^ and .Pd;2n/^ �

.Qd;2n/^^:

On the other hand, since Pd;2n and Qd;2n are closed, .Pd;2n/^^ D Pd;2n and

.Qd;2n/^^ D Qd;2n by the bipolar theorem (Proposition A.32). By the preceding we

have proved that Qd;2n D .Pd;2n/^ and Pd;2n D .Qd;2n/

^; or equivalently, that (19.24)and (19.25) are satisfied. ut


Let us call a truncated Rd-moment functional on Hd;m (as defined in Defini-tion 17.1) simply a moment functional. Our first main result in this section is thefollowing.

Theorem 19.14 Let f 2 Qd;m and

f .x/ DkX

jD1cj.yj � x/m with y1; : : : ; yk 2 Rd; c1 > 0; : : : ; ck > 0; k 2 N:

(19.27)Then there is a moment functional Lf onHd;m defined by

Lf . p/ WD 1

mŠf .@/p D Œ p; f � D

kXjD1

cjlyj. p/; p 2 Hd;m; (19.28)

with representing measure � D PkjD1 cjıyj : Each moment functional on Hd;m is of

the form Lf with f 2 Qd;m uniquely determined.

Proof Equation (19.16) gives (19.28). The latter means that Lf is a momentfunctional and � is a representing measure of Lf .

Conversely, let L be a moment functional on Hd;m. If L D 0, then L D Lf forf .x/ WD .0�x/m D 0. Now let L ¤ 0. By Theorem 17.2, L has a k-atomic representingmeasure � D Pk

jD1 cjıyj , where cj > 0 and k 2 N. Then f .x/ WD PkjD1 cj.yj � x/m

belongs to Qd;m and formula (19.16) implies that L D Lf .Suppose that Lf D Lg for f ; g 2 Qd;m. Then Œ p; f � D Œ p; g� by (19.28) and hence

Œ p; f�g� D 0 for all p 2 Hd;m. Because Œ�; �� is a scalar product on Hd;m, we concludethat f D g. ut

Obviously, c.y�x/m D .. mpc y/�x/m for c � 0. Therefore, since the polynomials in

Hd;m are homogeneous, the representation (19.27) of f 2 Qd;m in Theorem 19.14 andthe representing measure � of the truncated moment functional Lf are not unique.

There are several natural ways of normalizing these representations. Without lossof generality let us assume that f ¤ 0 and yj ¤ 0 for all j.

First, upon replacing yj by mpcj yj and allowing points of Rd, we can always

assume that cj D 1 for all j: Secondly, replacing yj by kyjk�1yj and cj by kyjkmcjand allowing coefficients cj > 0, we can assume that all points yj belong to the unitsphere Sd�1 of Rd. Further, thirdly, upon replacing yj by �yj if necessary, we canassume that all yj are in Sd�1C . Recall that Sd�1C denotes the set of all t D .t1; : : : ; td/of the unit sphere Sd�1 for which the first nonzero coordinate tj is positive. Thesethree normalizations mean to study the truncated moment problem for the vectorspace Hd;m on Rd, Sd�1; and Sd�1C ; respectively.

Since Sd�1C is a realization of the projective space Pd�1.R/ (see Sect. 17.2), thenormalization by Sd�1C refers to the truncated moment problem forHd;m onPd�1.R/.Hence the third normalization is most important among the three.


Let us retain the notation of Theorem 19.14. Then the Hankel matrix H.Lf / ofthe moment functional Lf has the entries h˛;ˇ; ˛; ˇ 2 N.d; n/; given by

h˛;ˇ D Lf .x˛Cˇ/ D Œx˛Cˇ; f � D

kXjD1

cj x˛Cˇ.yj/ D

kXjD1

cj y˛1Cˇ1j1 � � � y˛dCˇd

jd :

If s.y/ is the column vector .y˛/˛2N.d;n/ for y 2 Rd, then by Proposition 17.21,

H.Lf / DkX

jD1cj s.yj/ s.yj/

T : (19.29)

Our next theorem is a classical result of D. Hilbert (1909).

Theorem 19.15 Let d 2 N and n 2 N. Then the polynomial

hd;2n.x/ WD kxk2n � .x21 C � � � C x2d/n

is in Qd;2n, that is, there exist k 2 N and vectors y1; : : : ; yk 2 Rdnf0g such that

hd;2n.x/ DkX

jD1.yj � x/2n: (19.30)

The points yj can be chosen such that y1 D ky1ku, where u 2 Sd�1 is arbitrary.

Proof Let � denote the surface measure of the unit sphere Sd�1 and define

h.x/ WDZSd�1

.y � x/2nd�.y/; x 2 Rd:

We expand .y � x/2n by the multinomial theorem and conclude that h.x/ is ahomogeneous polynomial of degree 2n, that is, h 2 Hd;2n: Fix x 2 Rd. For y 2 Sd�1we abbreviate z1.y/ WD kxk�1.y � x/: Since jz1.y/j � 1, there is an orthogonaltransformation of variables y 7! z.y/ D .z1.y/; : : : ; zd.y// on Sd�1. Using theinvariance of � under an orthogonal change of variables and the relation y�x D z1kxkwe derive

h.x/ DZSd�1

.z1kxk/2nd�.z/ D kxk2nZSd�1

z2n1 d�.z/ D kxk2nc (19.31)

for some c > 0 depending on d; n. (For the explicit value of c, see Sect. 14.5.)On the other hand, we define a truncated Sd�1-moment functional on Hd;2n by

L. f / DZSd�1

f .y/ d�.y/; f 2 Hd;2n:


Obviously, if f D 0 on Sd�1 for some f 2 Hd;2n, then f D 0 on Rd and so f D 0.That is, the map f 7! f dSd�1 of Hd;2n is injective. Setting X D Sd�1 and E ŠHd;2n we are in the setup of Sect. 1.2 and L is a moment functional on E. Clearly,if L. f / D 0 for some f 2 EC Š Pos.Hd;2n; Sd�1/, then f D 0. That is, L is strictlyEC-positive. Therefore, by Theorem 1.30(ii), L has a k-atomic representing measure� DPk

jD1mjıtj which has u as an atom, say t1 D u. Then

L. f / DkX

jD1mj f .tj/ D

kXjD1

f .m1=2nj tj/: (19.32)

Specializing to f .y/ D .y �x/2n we get L. f / D h.x/ D ckxk2n DPkjD1.mj

1=2n tj �x/2nby (19.31) and (19.32). Setting yj D c�1=2n m1=2nj tj the last equation yields (19.30).By construction, y1ky1k�1 D t1 D u. ut

For many pairs .d; n/ explicit representations (19.30) of hd;2n are known byclassical formulas [Re1, Nat]. For instance, the following polynomial identitieshold:

12 h3;4 D 8x41 C 8x42 C 8x43C.x1 C x2 C x3/

4 C .x1 � x2 C x3/4 C .x1 C x2 � x3/

4 C .x1 � x2 � x3/4;

6 h4;4 D .x1 C x2/4 C .x1 � x2/

4 C .x1 C x3/4 C .x1 � x3/

4 C .x1 C x4/4 C .x1 � x4/

4C.x2 C x3/

4 C .x2 � x3/4 C .x2 C x4/

4 C .x2 � x4/4 C .x3 C x4/

4 C .x3 � x4/4;

60 h3;6 D 40x61 C 40x62 C 40x63C4.x1 C x2/

6 C 4.x1 � x2/6 C 4.x1 C x3/

6 C 4.x1 � x3/6C

.x1 C x2 C x3/6 C .x1 � x2 C x3/

6 C .x1 C x2 � x3/6 C .x1 � x2 � x3/

6:

As usual, let � D . @@x1/2 C � � � C . @

@xd/2 denote the Laplacian.

Theorem 19.16 Let d; n 2 N. There is a truncated Sd�1-moment functional onHd;2n defined by

L. p/ D 1

.2n/Š�np D Œ p; .x21 C � � � C x2d/

n�; p 2 Hd;2n: (19.33)

For each representation (19.30) of hd;2n.x/ 2 Qd;2n we have

L. p/ DkX

jD1p.yj/; p 2 Hd;2n: (19.34)


The functional L is strictly Pos.Hd;2n; Sd�1/-positive, that is, L. f / > 0 for eachpolynomial f 2 Pos.Hd;2n; Sd�1/; f ¤ 0.Proof Since hd;2n 2 Qd;2n by Theorem 19.15, it follows from Theorem 19.14 thatL � Lhd;2n is a truncated Sd�1-moment functional and (19.30) implies (19.34).

We show that L is strictly positive. Suppose that f 2 Pos.Hd;2n; Sd�1/; f ¤ 0.Then there is a u 2 Sd�1 such that f .u/ > 0. By Theorem 19.15, there exists arepresentation (19.30) such that y1 D ky1ku ¤ 0. Then, by (19.30),

L. f / DkX

jD1f .yj/ � f .y1/ D ky1k2nf .u/ > 0: ut

At the end of this section, we turn briefly to Carathéodory numbers.

Definition 19.17 For f 2 Qd;m the width w. f / of f is the smallest number k amongall possible representations (19.27) of f , where we set w. f / D 0 for f D 0.

For instance,w.h3;6/ D 11 andw.h3;8/ D 16, see [Re1, Theorems 9.28 and 9.37].Let L be a moment functional on Hm;d. Then, by Theorem 19.14, L D Lf for

some unique function f 2 Qd;m. We define w.L/ WD w. f /. Further, let C.L/ denotethe smallest number k for all k-atomic representing measures � 2 MC.Rd/ of L.

From Definition 18.7 we recall that the Carathéodory number

Cd;m WD C.Hd;m;Rd/

is the maximum of the numbers C.L/ for all moment functionals L on Hd;m.

Proposition 19.18 For each moment functional L on Hd;m we have w.L/ D C.L/:The number Cd;m is the largest width w.L/ for all moment functionals L onHd;m.

Proof By definition, C.L/ D w.L/ D 0 if L D 0. Hence we can assume that L ¤ 0.Throughout this proof, we abbreviate k WD C.L/ and r WD w.L/.

Let � be a k-atomic representing measure of L, say � DPkjD1 cjıyj , where cj >

0, yj 2 Rd for j D 1; : : : ; k. Then f .x/ WDPkjD1 cj.yj � x/m 2 Qd;m and

L. p/ DZ

p.x/d� DkX

jD1cjp.yj/ D Lf . p/ for p 2 Hd;m;

so that L D Lf . Hence r D w. f / � k.Conversely, let L be a moment functional. By Theorem 19.14, L D Lf for some

f 2 Qd;m. We can write f .x/ DPrjD1 cj.yj � x/m with yj 2 Rd; j D 1; : : : ;w. f / D r.

Then, by Theorem 19.14, � WD PrjD1 cjıyj is a representing measure for Lf D L

and � is k-atomic with k � r. Hence k � r. Thus we have proved that r D k.The second assertion follows at once from the equality w.L/ D C.L/: ut

As noted above, each moment functional of Hd;m is a truncated Sd�1-momentfunctional and Pos.Hd;m;R

d/ D Pos.Hd;m; Sd�1/. Hence, Theorem 18.43(ii),

19.4 Robinson’s Polynomial and Some Examples 483

applied with K D Sd�1, and (19.1) yield the following result: If the equality

Pos.Hd;2n;Rd/ D

XH2

d;n; (19.35)

that is, Pd;2n DP2d;2n; is satisfied, then Cd;2n D jNd;nj D

�nCd�1d�1

�:

Unfortunately, (19.35) holds only in very few cases. D. Hilbert [H1] has shownthat the equality (19.35) is fulfilled if and only if n D 1 or d D 2 or .d; n/ D .3; 2/.

It is not difficult to verify (see Exercises 19.7–19.9) that (19.35) holds if n D 1

or d D 2 and that (19.35) does not hold if d � 3; n � 3 or d � 4; n � 2.The remaining case .d; n/ D .3; 2/ is more difficult. A simple proof of the fact

that polynomials of Pos.H3;4;R3/ are sums of squares can be found in [CL].

Taking these results for granted we get

Cd;2 D d; C2;2n D nC 1; C3;4 D 6:

The general formula for Cd;2n is not yet known.

19.4 Robinson’s Polynomial and Some Examples

For typographical simplicity we write x; y; z instead of x0; x1; x2 in this section. Ouraim is to investigate the (homogeneous) Robinson polynomial

R.x; y; z/ WD x6 C y6 C z6 � x4y2 � x4z2 � x2y4 � y4z2 � x2z4 � y2z4 C 3x2y2z2:

Clearly, R 2 H3;6 and R is symmetric in all three variables x; y; z: It is convenient towrite R in two other forms:

R D x2.x2 � z2/2 C y2.y2 � z2/2 � .x2 � z2/.y2 � z2/.x2 C y2 � z2/ (19.36)

D .x2 C y2 � z2/.x2 � y2/2 C .x2 � z2/.y2 � z2/z2: (19.37)

Proposition 19.19

(i) R 2 Pos.R3/.(ii) Let .x; y; z/T 2 R3. Then we have R.x; y; z/ D 0 if and only if x D 0; y2 D z2

or y D 0; x2 D z2 or z D 0; x2 D z2 or x2 D y2 D z2.(iii) R is not a sum of squares in RŒx; y; z�.

Proof

(i) If .x2 � z2/.y2 � z2/.x2 C y2 � z2/ � 0, then it is obvious from (19.36) thatR.x; y; z/ � 0: Thus it remains to consider the case when

.x2 � z2/.y2 � z2/.x2 C y2 � z2/ � 0:


Then we have x2 C y2 � z2 � 0 and .x2 � z2/.y2 � z2/ � 0. But in thiscase (19.37) implies that R.x; y; z/ � 0: (Another way to prove (i) is to verifythe identity

.x2 C y2/R D x2z2.x2 � z2/2 C y2z2.y2 � z2/2 C .x2 � y2/2.x2 C y2 � z2/2:

Obviously, this implies that R � 0 on R3.)(ii) The if part is clear from (19.36). Conversely, suppose that R.x; y; z/ D 0. If

x D 0, then (19.36) yields y2.y2 � z2/2 C z2.y2 � z2/2 D 0 and hence y2 D z2.Similarly, since R is symmetric, y D 0 implies x2 D z2 and z D 0 impliesx2 D y2.

Suppose now that x ¤ 0; y ¤ 0; z ¤ 0. Assume to the contrary that x2 Dy2 D z2 does not hold. By symmetry we can assume without loss of generalitythat x2 > z2. Then .x2�z2/.x2Cy2�z2/ > 0: Therefore, since x2.x2�z2/2 > 0,the last term in (19.36) must be negative, which implies that z2 < y2. Then itfollows from (19.37) that R.x; y; z/ > 0, which is a contradiction.

(iii) Assume to the contrary that R D Pj f2j . Clearly, deg. fj/ � 3. We write fj as

fj D xpj C qj, where pj and qj contain only even powers of x. Then

R DX

j.x2p2j C 2xpjqj C q2j /: (19.38)

Comparing the coefficients of x6 we conclude that

1 DX

jpj.1; 0; 0/

2: (19.39)

Since R has only even powers of x, the odd powers of x on the right-hand sideof (19.38) vanish. Thus,

Pj 2xpjqj D 0: Then (19.38) implies that R � x2p2j

on R3 for all j. Setting y D x and writing pj.x; x; z/ D aix2 C bixzC ciz2 thisyields

R.x; x; z/ D z2.z2 � x2/2 � x2pj.x; x; z/2 D x2.aix

2 C bixzC ciz2/2:

Choosing x large, it follows that ai D 0. Then z2.z2 � x2/2 � x2.bix C ciz/2:Hence bix C ciz D 0 if x ˙ z D 0; x ¤ 0. Therefore, bi D ci D 0; so thatpj.x; x; z/ D 0. Similarly, p.x;�x; z/ D 0. These relations imply that pj.x; y; z/is divisible by x� y and xC y. Replacing y by z in the preceding, pj is divisibleby x � z and xC z. Thus,

pj.x; y; z/ D .x2 � y2/.x � z2/gj:

Since deg. pj/ � 2, gj � 0 and pj.1; 0; 0/ D 0 for all j which contra-dicts (19.39).

An elegant and short proof of assertion (iii) can be given by using a basicresult from the theory of cubics, see Example 19.36 below. ut

19.4 Robinson’s Polynomial and Some Examples 485

As discussed in Sect. 17.2, the subset S2C of the unit sphere S2 can be consideredas a realization of the projective space P2.R/. From the description of the zeroset of R given in Proposition 19.19(ii) it follows that R has exactly 10 zeros in theprojective space P2.R/. The corresponding points of S2C are

t1 D 1p3.1; 1; 1/; t2 D 1p

3.1; 1;�1/; t3 D 1p

3.1;�1; 1/; t4 D 1p

3.1;�1;�1/;

t5 D 1p2.1; 0; 1/; t6 D 1p

2.1; 0;�1/; t7 D 1p

2.1; 1; 0/; t8 D 1p

2.1;�1; 0/;

t9 D 1p2.0; 1; 1/; t10 D 1p

2.0; 1;�1/:

Proposition 19.20 Let p.x; y; z/ 2 H3;6: Then we have p 2 Pos.R3/ and p.tj/ D 0

for j D 1; 2; : : : ; 9 if and only if p 2 RC � RCRC � h210, where

h10.x; y; z/ D y3 � z3 C .z � y/x2 D .y � z/.y2 C yzC z2 � x2/:

Proof The if part is easily verified. It suffices to prove the only if part.Suppose that p 2 Pos.R3/ and p.tj/ D 0 for j D 1; 2; : : : ; 9: Write

p.x; y; z/ D6X

j;k;lD0ajklx

jykzl:

Since p � 0 on R3, each zero tj is a local minimum of p, so all first-order partialderivatives of p at tj vanish as well (see also Lemma 19.26 below). Thus

p.tj/ D @p

@x.tj/ D @p

@y.tj/ D @p

@z.tj/ D 0 for j D 1; : : : ; 9: (19.40)

These are 36 homogeneous linear equations for the 28 variables ajkl. It can beshown (for instance, by using a computer program such as Mathematica) that thecorresponding matrix has rank 26. Hence the dimension of the solution space is 2.Since R and h210 are in Pos.R3/ and vanish at t1; : : : ; t9, the coefficient vectors of Rand h210 are solutions of (19.40). Hence p D �1R C �2h210 with �1; �2 2 R: SinceR.t10/ D 0 and h10.t10/2 > 0, we get p.t10/ D �2h10.t10/2 � 0, so that �2 � 0. Fort D .2; 1; 1/ we have h10.t/ D 0 and R.t/ ¤ 0. Then, p.t/ D �1R.t/ � 0 and hence�1 � 0. utRemark 19.21 Let i 2 f1; : : : ; 10g:The assertion of Proposition 19.20 remains validif t10 is replaced by ti: There is a polynomial hi 2 H3;3 such that p 2 Pos.H3;6;R

3/

vanish on tj for j D 1; : : : ; 10; j ¤ i; if and only if p 2 RC � RCRC � h2i : We have

h1 D x.y2 C z2 � x2/C y.z2 C x2 � y2/C z.x2 C y2 � z2/ � xyz:


All other polynomials hi can be obtained by symmetry from h10 and h1; for instance,h5.x; y; z/ D h10.y; x;�z/; h2.x; y; z/ D h1.x; y;�z/ etc. ıCorollary 19.22 Let p 2 Pos.H3;6;R

3/. The following are equivalent:

(i) p.tj/ D 0 for j D 1; : : : ; 10:(ii) p D �R for some number � � 0:

(iii) .�R � p/ 2 Pos.R3/ for some number � � 0:Proof

(i)!(ii) By Proposition 19.20, (i) implies p D �1RC�2h210 with �1 � 0, �2 � 0.Since h10.t10/ ¤ 0 and R.t10/ D 0, we obtain �2 D 0; so that p D �1R:

(ii)!(iii) is trivial by setting � D �.(iii)!(i) Since 0 � p � �R on R3 and R.tj/ D 0 for j D 1; : : : ; 10, p vanishes

at t1; : : : ; t10 as well. utFrom Corollary 19.22 (iii)!(ii) it follows that the Robinson polynomial R spans

an extreme ray of the cone Pos.H3;6;R3/:

Now we turn to the truncated moment problem on P2.R/ Š S2C and set

N WD f˛ D .˛0; ˛1; ˛2/ 2 N30 W ˛0 C ˛1 C ˛2 D 6g:

Then Lin fx˛ W ˛ 2 Ng is the vector space H3;6 of 3-forms of degree 6.We fix a point t0 2 S2C such that t0 ¤ tj for j D 1; : : : ; 10. Then R.t0/ ¤ 0. Put

� D10XjD1

mjıtj ; � D � Cm0ıt0 D10XjD0

miıti ; where mj � 0; j D 0; : : : ; 10;

and let L� and L� denote the corresponding truncated moment functionals on theprojective space P2.R/ Š S2C. The following simple fact will be used several times.

Lemma 19.23 Let p 2 H3;6. If p.x/ � 0 on S2C, then p.x/ � 0 on R3.

Proof The equalities R3 D [c2R c S2C and p.cx/ D c6p.x/ give the assertion. utIn the following proofs we use the projective versions of the notions and results.

Proposition 19.24 L� is a determinate truncated moment functional on P2.R/ ŠS2C and � is a maximal mass measure. Further, if mj > 0 for j D 1; : : : ; 10, then

NC.L�; S2C/ D RC � R and VC.L� ; S2C/ D supp � D ft1; : : : ; t10g: (19.41)

Proof By straightforward computations we verify that the polynomials

q1 D x2y.xC y/z.xC z/; q2 D x2.x � y/yz.x � z/;

q3 D x2y.xC y/z.x � z/; q4 D x2y.x � y/z.xC z/;

19.5 Zeros of Positive Homogeneous Polynomials 487

q5 D x2.x2 � y2/z.xC z/; q6 D x2.x2 � y2/z.x � z/;

q7 D x2.x � y/.x2 � z2/; q8 D x2.xC y/.x2 � z2/;

q9 D .x2 � y2/.x2 � z2/.y � z/2; q10 D .x2 � y2/.x2 � z2/.yC z/2

of H3;6 satisfy qj.ti/ D 0 for j ¤ i and qj.tj/ ¤ 0, where i; j D 1 : : : ; 10: From thisit follows that the points t1; : : : ; t10 obey property .SP/H3;6 , see Definition 18.11.Hence L� is determinate by Proposition 18.12. In particular, � is maximal mass.

We abbreviate NC WD NC.L� ; S2C/ and prove that NC D RC � R: Since mj > 0

for j D 1; : : : ; 10, NC consists of p 2 Pos.H3;6; S2C/ that vanish at t1; : : : ; t10,see (18.11). Hence R 2 NC. Conversely, let p 2 NC. Then p � 0 on S2C. Hencep � 0 on R3 by Lemma 19.23, so p 2 RC � R by Corollary 19.22 (i)!(ii). Thus,NC D RC � R:

Since ft1; : : : ; t10g is the zero set of R on the projective space, the second equalityof (19.41) follows at once from the first. utProposition 19.25 Suppose that mj > 0 for j D 0; 1; : : : ; 10. Then the measure� isordered maximal mass. If h.t0/ ¤ 0 and R.t0/ > 0 (for instance, if t0 D .1; 0; 0/),then � is not maximal mass.

Proof In this proof we abbreviate L D L�.Put p WD R.t0/�1R: Then p � 0 on S2C, p.t0/ D 1, and p.tj/ D 0 for j D 1; : : : ; 10:

Hence L.t0/ D m0 by Proposition 18.31(iii). Since ��m0ıt0 D � by definition and� is maximal mass by Proposition 19.24, � is ordered maximal mass.

Let f 2 Pos.H3;6; S2C/ and suppose that L. f / D 0. Then f � 0 on S2C and hencef � 0 on R3 by Lemma 19.23. From L. f / D 0 and mj > 0 we obtain f .tj/ D 0

for j D 0; 1; : : : ; 10. Therefore, by Corollary 19.22, f D �R with � � 0. Sincef .t0/ D 0 and R.t0/ ¤ 0, f D 0. This shows that L is strictly Pos.H3;6; S2C/-positive. Therefore, by Proposition 18.28(iii), the infimum in (18.23) is attained foreach point x.

Assume to the contrary that � is maximal mass. Then Theorem 18.33(ii) appliesto the compact set K D P2.R/ Š S2C; so there exists a polynomial p 2 H3;6 suchthat p � 0 on S2C, hence p � 0 on R3 by Lemma 19.23, and p.t10/ D 1, p.tj/ D 0 forj D 0; : : : ; 9. From Proposition 19.20 we obtain p D �1RC�2h210 with �1; �2 2 RC:Since R.t0/ > 0 and h10.t0/ ¤ 0 by assumption, p.t0/ D �1R.t0/ C �2h.t0/2 D 0

implies that �1 D �2 D 0. Thus, p D 0 and p.t10/ D 0: This is a contradiction. ut

19.5 Zeros of Positive Homogeneous Polynomials

Zeros of positive polynomials play a crucial role for the truncated moment problem.They appeared in the definition of the subcone NC.L;K/ (see (18.11)) and in thestudy of maximal masses (see Definition 18.32). In this section, we develop somebasic results on zeros of positive polynomials of H3;2n.


We begin with two simple preliminary lemmas.

Lemma 19.26 Let U be an open subset of Rd, u 2 U, and f 2 RdŒ x �. Supposethat f .u/ D 0 and f .x/ � 0 for all x 2 U. Then we have @f

@xi.u/ D 0 for i D 1; : : : ; d

and the Hessian matrix�@2f@xi@xj

.u/�di;jD1 is positive semidefinite.

Proof Upon translation we can assume that u D 0. Let t D .t1; : : : ; td/ 2 U. Sincef .0/ D 0, the Taylor expansion of f at 0 is

f .t/ DdX

iD1ti@f

@xi.0/C

dXi;jD1

titj@2f

@[email protected]/C

dXi;j;kD1

titjtl@3f

@xi@[email protected]/C � � � :

We set tj D 0 for j ¤ i and consider small jtij. Since f .t/ � 0 on U, this expansionimplies that @f

@xi.u/ D 0. Thus the linear part of the Taylor expansion disappears.

Using again that f .t/ � 0 on U and taking all jtij small we conclude that thequadratic part is nonnegative. Hence the Hessian is a positive semidefinite matrix.

utLemma 19.27 Let f 2 RdŒ x �. Suppose that f is irreducible inRdŒ x �, but reduciblein CdŒ x �. Then there exist polynomials g1; g2 2 RdŒ x � such that f D �.g21 C g22/,where � D 1 or � D �1.Proof Since f is reducible in CdŒ x �, there is a nontrivial factorization f D qp withp; q 2 CdŒ x �. Taking complex conjugates and using that f 2 RdŒ x � we get f D q p.Hence f � f D .qq/ � . pp/: Since f is irreducible in RdŒ x � and qq; pp 2 RdŒ x �, theuniqueness of irreducible factorizations implies that f D a.q q/ for some nonzeroreal number a:We write

pjaj q D g1C ig2 with g1; g2 2 RdŒ x � and set � D ajaj�1.Then f D a .q q/ D �.g21 C g22/. ut

Next we collect some facts on plane curves. We will need only a few elementarynotions and some basic results stated as Lemmas 19.28, 19.29, and 19.38.

Let f 2 H3;m: Then the set

ZP. f / D ft 2 P2.R/ W f .t/ D 0g

is called the real projective curve associated with f , or briefly, the curve f D 0. Notethat the equation f .t/ D 0 is well-defined for t 2 P2.R/, since f is homogeneous.

A point t of the curve ZP. f / is called singular if @[email protected]/ D @f

@x2.t/ D @f

@x3.t/ D 0:

Let t D Œt1 W t2 W t3� 2 ZP. f /. If f � 0 in some neighbourhood U R3 of.t1; t2; t3/ 2 R3, Lemma 19.26 implies that t is a singular point of the curve ZP. f /.

Lemma 19.28 If f 2 H3;m is irreducible in C3Œ x �, then the curve ZP. f / has atmost 1

2m.m � 1/ singular points.

Proof See e.g. [Wr, II, Theorem 4.4]. utLet f 2 H3;m and g 2 H3;n. For each point t 2 ZP. f / \ ZP.g/ the intersection

multiplicity It. f ; g/ 2 N of the curves f D 0 and g D 0 at t is defined in

19.5 Zeros of Positive Homogeneous Polynomials 489

[Wr, III, Section 2.2] or in [Fn, Chapter 3, Section 3.3]. We do not restate theprecise definition here. In what follows we use only the fact that It. f ; g/ � 2 ift 2 ZP. f / \ZP.g/ is a singular point of one of the curves f D 0 or g D 0.

Also, we will need Bezout’s theorem. For complex projective curves we evenhave equality in (19.42), but for our applications the following simple version issufficient. The symbol jZj denotes the number of points of a set Z.

Lemma 19.29 Suppose that the polynomials f 2 H3;n and g 2 H3;m are relativelyprime in R3Œ x �, that is, f and g have no common factor of positive degree. ThenjZP. f / \ZP.g/j � nm: More precisely,

Xt2ZP. f /\ZP.g/

It. f ; g/ � nm: (19.42)

Proof See e.g. [Wr, p. 59] or [Fn, p. 57]. utHaving finished the preparations we turn to zeros of positive polynomials.

Proposition 19.30 Let f 2 H3;2n be irreducible in R3Œ x �. If f 2 Pos.R3/, then

jZP. f /j � ˛.n/ WD max.n2; .2n � 1/.n � 1//: (19.43)

Proof First assume that f is also irreducible in C3Œ x �. Since f 2 Pos.R3/, byLemma 19.26 all zeros of f are singular points of the curve f D 0. Since f isirreducible in C3Œ x �, this curve has at most 1

2.2n � 1/.2n � 2/ D .2n � 1/.n � 1/

singular points by Lemma 19.28. Therefore, jZP. f /j � .2n� 1/.n� 1/:Now let f be reducible in C3Œ x �: Then Lemma 19.27 applies. Since f � 0 on R3,

we have f D g21 C g22 with g1; g2 2 R3Œ x �. Clearly, g1; g2 are also homogeneous,ZP. f / D ZP.g1/ \ ZP.g2/, and deg.gj/ � n for j D 1; 2. We verify that g1 andg2 are relatively prime in C3Œ x �. Indeed, if h 2 C3Œ x � is a common factor of g1and g2, then h � h is a common factor of g21 and g22 and so of f : Therefore, h � h, andhence h, is constant, because f is irreducible in R3Œ x �. That is, the assumptions ofBezout’s theorem (Lemma 19.29) are satisfied, so the real projective curves definedby g1 D 0 and g2 D 0 intersect in at most n � n points. Thus, jZP. f /j � n2: utLemma 19.31 The function ˛.n/ onN defined by (19.43) is subadditive, that is,

˛.n1/C ˛.n2/C � � � C ˛.nr/ � ˛.n1 C � � � C nr/ for n1; : : : ; nr 2 N:

(19.44)

Proof Note that ˛.n/n is monotone increasing on N. Therefore, ˛.nj/nj� ˛.n1Cn2/

n1Cn2for

j D 1; 2 and hence

˛.n1/C ˛.n2/ � n1˛.n1 C n2/

n1 C n2C n2˛.n1 C n2/

n1 C n2D ˛.n1 C n2/:

Formula (19.44) follows by induction on r. ut


Clearly, ˛.1/ D 1; ˛.2/ D 4; ˛.3/ D 10, and ˛.n/ D .2n � 1/.n� 1/ for n � 3.Note that the Robinson polynomial R 2 H3;6 has exactly ˛.3/ D 10 zeros in P2.R/.

Let us call a polynomial f 2 RdŒ x � indefinite if neither f nor �f is in Pos.Rd/,that is, there exist points t1; t2 2 Rd such that f .t1/ < 0 and f .t2/ > 0.

Our main result in this section is the following theorem.

Theorem 19.32 Let n 2 N and f 2 H3;2n: Suppose that f .x/ � 0 on R3 andjZP. f /j > ˛.n/: Then jZP. f /j is infinite and there are polynomials p 2 H3;2n1 ,q 2 H3;n2 such that f D pq2, where n1 C n2 D n, p 2 Pos.R3/, jZP. p/j <1, q isindefinite, and jZP.q/j is infinite. Moreover, q is a product of indefinite irreduciblepolynomials in R3Œ x � and p and q are relatively prime in R3Œ x �: (It is possible thatp is a positive real constant; in this case n1 D 0 and we set H3;0 WD R.)

Proof Since the assertion trivially holds for f D 0, we can assume that f ¤ 0. Letf D f1 � � � fr be a factorization of f as a product of irreducible factors in R3Œ x �: Sincef is homogeneous, so are all factors fj. Set mj D deg. fj/.

Assume first that no factor fj is indefinite. Then mj D 2nj is even. Multiplying by�1 if necessary, we can suppose that fj 2 Pos.R3/2nj . Then, by (19.43) and (19.44),

jZP. f /j �X

jjZP. fj/j �

Xj˛.nj/ � ˛.n1 C � � � C nr/ D ˛.n/;

which contradicts the assumption jZP. f /j > ˛.n/:As shown in the preceding paragraph, there is at least one indefinite factor, say f1.

It suffices to show that jZP. f1/j is infinite and that f 21 divides f . Indeed, then f D f 21 hwith h 2 H3;2k; k < n. Clearly, h � 0 on R3. If ZP.h/ is finite, we are finished; ifnot, we proceed by induction.

Set g D f2 � � � fr: Then f D f1g. Since f1 is indefinite, there are t1; t2 2 R3 suchthat f1.t1/ < 0 and f1.t2/ > 0. After affine changes of coordinates we can assumethat t1 D .1; 0; a/, t2 D .1; 0; b/, a < b. By continuity, there is a ı > 0 such thatf1.1; u; a/ < 0 and f1.1; u; b/ > 0 for juj < ı: Fix u 2 .�ı; ı/: Since f1.1; u; y/is a nonconstant polynomial in y, by the intermediate value theorem there exists acu 2 .a; b/ such that f1.1; u; cu/ D 0 and f1.1; u; y/ < 0 for y < cu, f1.1; u; y/ > 0

for y > cu in a neighbourhood of cu. From f .1; u; y/ D f1.1; u; y/g.1; u; y/ � 0 weget g.1; u; y/ � 0 if y < uc, g.1; u; y/ � 0 if y > uc in a neighbourhood of cu. Henceg.1; u; cu/ D 0; so that .1 W u W cu/ 2 ZP. f1/ \ ZP.g/. Thus, ZP. f1/ \ ZP.g/,hence ZP. f1/, is infinite. Clearly, f1 and g are not relatively prime, because thiswould imply that their curves have only finitely many common points by Bezout’stheorem. Since f1 is irreducible, f1 divides g. Because f D f1g, then f 21 divides f .

By construction, q is a product of irreducible indefinite factors fi. Since ZP. p/ isfinite and each ZP. fi/ is infinite, p and q are relatively prime in R3Œ x �: ut

The reasoning concerning f1 in the preceding proof yields the following.

Corollary 19.33 For each indefinite polynomial f 2 H3;m the set ZP. f / is infinite.

Remark 19.34 As noted in [CLR, p. 13], Theorem 19.32 remains valid if ˛.n/ isreplaced by O .n/ WD 3

2n.n�1/C1: This stronger statement is based on deep results

19.6 Applications to the Truncated Moment Problem on R2 491

of I.G. Petrovskii and O.A. Oleinik on ovals of plane curves [PO]. Let ˇ.n/ denotethe maximum of jZP. f /j, where f 2 H3;2n and ZP. f / is finite. Then

n2 � ˇ.n/ � O .n/ D 3

2n.n � 1/C 1: ı

19.6 Applications to the Truncated Moment Problem on R2

In this section, we suppose that t1; : : : ; tk are pairwise distinct points of R2,c1; : : : ; ck are positive numbers and L denotes the truncated moment functional onR2Œ x �2n; where n D 2; 3, given by the k-atomic measure � DPk

jD1 cjıtj :

L. f / DZ

f d� �kX

jD1cj f .tj/; f 2 R2Œ x �2n: (19.45)

Then the vector space NL and the cone NC.L/ from Definitions 17.16 and 18.13 are

NL D fp 2 R2Œ x �n W p.t1/ D � � � D p.tk/ D 0 g; (19.46)

NC.L/ D fp 2 Pos.R2/2n W p.t1/ D � � � D p.tk/ D 0 g; (19.47)

and VL and VC.L/ are the real algebraic sets in R2 defined by NL and NC.L/,respectively.

Our aim is to illustrate the usefulness of Theorem 19.32 by developing someresults on VL and VC.L/ for n D 2; 3. Since Theorem 19.32 deals with forms, wehave to switch between polynomials of R2Œ x �2n and forms of H3;2n.

Let q 2 R2Œ x �m be of degree m and let p 2 H3;m. We define the homogenizationqh 2 H3;m and the dehomogenization Op 2 R2Œ x �m by

qh.x1; x2; x3/ WD xm1 q

�x2x1;x3x1

�and Op.x1; x2/ WD p.1; x1; x2/: (19.48)

Then bqh D q and deg.qh/ D deg.q/ D m. Clearly, deg.Op/ � m, but it may happenthat deg.Op/ < m; for instance, if p.x1; x2; x3/ D x21 and m D 2, then Op D 1: Ifdeg.Op/ D m, then .Op/h D p: If pj 2 H3;mj for j D 1; 2, then p1p2 2 H3;m1Cm2 andbp1p2 D bp1bp2. For even m, it is obvious that q 2 Pos.R2/ if and only if qh 2 Pos.R3/.

The zero set Z.q/ in R2 is a subset of the zero set ZP.qh/ in P2.R/. In general,ZP.qh/ is larger than Z.q/, because it contains possible zeros of q at “infinity”.

We will use the following version of Bezout’s theorem for real curves in R2

which follows from Lemma 19.29: Let f ; g 2 R2Œ x �; deg. f / D m1, and deg.g/ Dm2. If f and g are relatively prime in R2Œ x �, then jZ. f /\ Z.g/j � m1m2 and

Xt2Z. f /\Z.g/ It. f ; g/ � m1m2: (19.49)


By a quadratic or a cubic we mean a real curve Z. p/ in R2 for some polynomialp 2 R2Œ x � of degree 2 or 3, respectively.

Proposition 19.35 Let n D 2 and k D 5. Suppose that t1; : : : ; t5 are distinct pointsof R2, no three are collinear. There exists a unique, up to a constant multiple,polynomial f 2 R2Œ x � of degree 2 such that t1; : : : ; t5 lie on the quadratic Z. f /.Then

NC.L/ D RC � f 2 and VC.L/ D Z. f /: (19.50)

Proof The existence and uniqueness of f 2 R2Œ x �2 satisfying t1; : : : ; t5 2 Z. f / is aclassical result in curve theory, see e.g. [Bx, Theorem 5.10].

Let g 2 NC.L/, g ¤ 0. Then, by (19.47), g 2 Pos.R2/ and g vanishes att1; : : : ; t5. If deg.g/ D 2, then g is a sum of squares of linear polynomials andall tj are collinear, a contradiction. Hence deg.g/ D 4, so that gh 2 H3;4. Wehave gh 2 Pos.R3/ and jZP.gh/j � jZ.g/j � 5 > 4 D ˛.2/. Therefore, byTheorem 19.32, we can write gh D pq2, where q is indefinite, p 2 Pos.R3/ andZP. p/ is finite. Then g D Op Oq2:

Since q is indefinite, q is not constant. Assume that deg.q/ D 1. Then p 2 H3;2

and deg.Oq/ � 1. Thus, Oq ¤ 0 (by g ¤ 0) is either constant or Oq D 0 is a line.Since g.tj/ D Op.tj/Oq.tj/2 D 0 for j D 1; : : : ; 5 and no three points of the tj areon a line, Op vanishes on at least three points, so that jZP.Op/j � 3 > ˛.1/ D 1:

From Theorem 19.32, applied to p, it follows that ZP. p/ is infinite, which is acontradiction.

Thus, q 2 H3;2 and p is a nonzero constant. Then gh is a multiple of q2, so g is amultiple of Oq2: Since g 2 NC.L/ vanishes on t1; : : : ; t5, so does Oq2 and hence Oq. Theuniqueness assertion concerning f implies that Oq D cf for some c 2 R. Therefore,g 2 RC � f 2: This proves that NC.L/ RC � f 2: The converse inclusion and theequality VC.L/ D Z. f / are obvious. ut

The case of cubics is more subtle. In the following discussion we use someclassical results on cubics (see e.g. [Bx]) and we state some facts without proof.

Case 1: k D 8.

Suppose Z8 WD ft1; : : : ; t8g is set of pairwise distinct points of R2 such that nofour of them are collinear and no seven lie on a quadratic. Then NL is the vectorspace of polynomials p 2 R2Œ x �3 of degree 3 vanishing on Z8 and this space hasdimension 2. There is a point t9 in the projective zero set of NL which has thefollowing property: For each t10 2 R2, different from t1; : : : ; t9, there is a unique, upto a constant, cubic which contains the points t1; : : : ; t8; t10: Note t9 may or may notbe in the set Z8. (All preceding results follow from Theorems 13.4, 13.6, and 13.7in [Bx].)

Assume that t9 2 R2 and t9 … Z8: Let the cubics f ; g form a basis of NL: Then

Z. f /\Z.g/ D Z9 WD ft1; : : : ; t9g; (19.51)


that is, both cubics intersect in exactly 9 points of R2, and VL D Z. f /\Z.g/ D Z9.It is not difficult to verify that (19.51) implies that f and g are relatively prime.

As proved by B. Reznick [Re4, Theorem 3.4 and 4.1], there exists a polynomialp 2 Pos.R2/ such that p.tj/ D 0 for j D 1; : : : ; 8 and p.t9/ > 0. Then p 2 NC.L/and t9 … VC.L/. From Z8 D supp� VC.L/ VL D Z9 we get VC.L/ D Z8:Thus,

VC.L/ D Z8 ¨ Z9 D VL: (19.52)

We illustrate this with an example.

Example 19.36 The cubics f D x1.1 � x21/; g D x2.1 � x22/ satisfy (19.51),where the points t1; : : : ; t8 are .˙1;˙1/; .˙1; 0/; .0;˙1/ and t9 D .0; 0/. Thedehomogenized Robinson polynomial p WD OR is in Pos.R2/, vanishes at t1; : : : ; t8,and satisfies p.t9/ > 0. Hence (19.52) holds.

By Proposition 19.19(iii), Robinson’s polynomial R 2 H3;6 is not a sum ofsquares. The preceding yields a very short proof of this fact. Indeed, assume thatR D P

j f2j with fj 2 R3Œ x �. Then OR D P

j.Ofj/2 and Ofj 2 R2Œ x �3: Clearly, each

Ofj vanish at all zeros of OR. Therefore, since OR vanishes at Z8, Ofj 2 NL; so thatOfj 2 Lin f f ; gg. Then, (19.51) implies that Ofj.t9/ D 0 for all j: Hence OR.t9/ D 0.But OR.t9/ D R.1; 0; 0/ D 1, a contradiction. ı

Case 2: k D 9.

Retain the assumptions and the notation of Case 1. Since NL is spanned byf ; g, (19.51) yields VL D Z9. From Z9 D supp� VC.L/ VL we getZ9 D VC.L/ D VL.

Case 3: k � 10.

There exists a truncated moment functional L such that VC.L/ consists of 10points, see (19.41). (This example is on P2.R/, but by a linear transformation weobtain such a measure on R2, see Exercise 19.12.)

In the next proposition the real algebraic set VC.L/ is a cubic.

Proposition 19.37 Suppose that k � 10. Let tj; j D 1; : : : ; k; be pairwise distinctpoints in R2 of a cubic Z. f /, f 2 R2Œ x �3: Suppose that no four of the pointst1; : : : ; t8 are collinear and no seven of them are on a quadratic. Let � DPk

jD1 cjıtjbe a k-atomic measure and let L be its moment functional defined by (19.45). Then,

NC.L/ D RC � f 2 and VC.L/ D Z. f /:

Proof As in the proof of Proposition 19.35 it suffices to show that NC.L/ RC �f 2:Let g 2 NC.L/; g ¤ 0. Then g 2 Pos.R2/, so that gh 2 Pos.R3/, and g.tj/ D 0

for j D 1; : : : ; k: By Lemma 19.26, each tj is a singular point of Z.g/, so thatItj. f ; g/ � 2. Therefore,

Pj Itj. f ; g/ � k � 2 � 20 > 18 � deg.g/ deg. f /. Hence

Bezout’s theorem (19.49) implies that f and g are not relatively prime.


Let r be a nonconstant common factor of f and g, say f D ru and g D rv.We show that Z.gh/ is infinite. Since gh D rhvh, by Corollary 19.33 it suffices toshow that rh is indefinite. This is obvious if deg.r/ is odd. Let deg.r/ D 2. Thendeg.u/ D 1. We have f .tj/ D r.tj/u.tj/ D 0 for j D 1; : : : ; 9. Since no four oft1; : : : ; t9 lie on the line u D 0, r vanishes at 6 points ti. The polynomial r is not inPos.R2/, since then rh 2 Pos.R3/ and hence 6 � jZ.r/j � jZP.rh/j � ˛.2/ D 4

by Theorem 19.32, a contradiction. Similarly, �r is not in Pos.R2/. Thus, r, andhence rh, is indefinite.

Since ZP.gh/ is infinite, by Theorem 19.32 there is a factorization gh D q2p,where q is indefinite, p 2 Pos.R3/ and ZP. p/ is finite. Note that q is not constant,since g ¤ 0 and hence gh ¤ 0. Further, g D OpOq2: Thus, Oq ¤ 0, because g ¤ 0.

Case I: deg.q/ D 1:Then deg. p/ � 4, so that p 2 H3;2k for k 2 f0; 1; 2g: Since p 2 Pos.R3/ and

ZP. p/ is finite, we conclude from Theorem 19.32 that jZ.Op/j � jZP. p/j � ˛.2/ D4: We have g.tj/ D Op.tj/Oq.tj/2 D 0 for j D 1; : : : ; 8: Therefore, at least four of thepoints t1; : : : ; t8 lie on the curve Oq D 0. Since deg.Oq/ � deg.q/ D 1; Oq is either anonzero constant or Oq D 0 is a line and we obtain a contradiction to our assumption.

Case II: deg.q/ D 2:Then deg. p/ � 2, so that jZ.Op/j � ˛.1/ D 1: Hence, since g.tj/ D Op.tj/Oq.tj/2 D

0 for j D 1; : : : ; 8; seven of the points t1; : : : ; t8 lie on the curve Oq D 0. Sincedeg.Oq/ � 2, Oq is a nonzero constant or Oq D 0 is a line or a quadratic, again acontradiction.

From the preceding two cases it follows that deg.q/ D 3 and p is constant.Therefore deg.Oq/ D 3. Since g D OpOq2, Oq vanishes at all 10 points t1; : : : ; t10. Thesepoints determine the cubic uniquely [Bx, Theorem 13.7], so Oq is a constant multipleof f . Hence g D OpOq2 is a nonnegative multiple of f 2, that is, g 2 RC � f 2: ut

The next lemma is Chasles’ theorem. In the literature this result or its general-ization to curves of higher degrees is also called the Cayley–Bacharach theorem.

Lemma 19.38 Let t1; : : : ; t9 be pairwise distinct points of R2 and suppose thatf ; g 2 R2Œ x �3 define two cubics satisfying Z. f / \ Z.g/ D ft1; : : : ; t9g. Leth 2 R2Œ x �3. If the points t1; : : : ; t8 lie on the cubic Z.h/, then also t9 2 Z.h/.

Proof [EGH, Theorem CB3], see also [SR, Chapter V, Section 1.1] and [RRS].From the assumptions and Bezout’s theorem one easily derives that no four of thepoints tj are collinear and no seven are on a quadratic. Using this fact the assertionof Lemma 19.38 follows from [Bx, Theorem 13.7] as well. ut

Recall that a positive functional on R2Œ x � that is not a moment functional wasgiven in Proposition 13.5. Now we use the preceding setup to construct anotherpositive functional on R2Œ x �6 that is not a truncated moment functional. Let

t1 D .1; 1/; t2 D .1;�1/; t3 D .�1; 1/; t4 D .�1;�1/; (19.53)

t5 D .0; 1/; t6 D .0;�1/; t7 D .1; 0/; t8 D .�1; 0/; t9 D .0; 0/: (19.54)


The first 8 points t1; : : : ; t8 are zeros of the dehomogenized Robinson polynomial

OR.x1; x2/ WD R.1; x1; x2/ D x61 C x62 C 1 � x41.x22 C 1/ � x42.x

21 C 1/ � x41 � x42 C 3x21x22

and OR.t9/ D 1. For c 2 R, we define Lc DP8jD1 ltj � clt9 .

Proposition 19.39 Let 0 < c � 45. Then Lc is a positive functional onR2Œ x �6 (that

is, Lc. p2/ � 0 for p 2 R2Œ x �3) which is not a truncated moment functional.

Proof Recall that t1; : : : ; t9 are the points satisfying (19.36) for the two cubics f Dx1.1 � x21/ and g D x2.1 � x22/ from Example 19.36. Hence Lemma 19.38 applies.

First we show that the point evaluations lt1 ; : : : ; lt9 are linearly dependent onR2Œ x �3. Assume the contrary and let L be the linear span of these functionals. Thenthere is a linear functional F on L such that F.ltj/ D 0, j D 1; : : : ; 8; and F.lt9 / D 1.Since L .R2Œ x �3/�, the functional F is given by some polynomial p 2 R2Œ x �3,that is, F.l/ D l. p/ for l 2 L. Then F.ltj/ D p.tj/. Hence t1; : : : ; t8 2 Z. p/.Since t1; : : : ; t8 do not lie on a quadratic, deg. p/ D 3 and t1; : : : ; t8 are on the cubicZ. p/. Hence, by Chasles’ theorem (Lemma 19.38), p.t9/ D 0 which contradictsp.t9/ D F.lt9 / D 1:

Since the functionals lt1 ; : : : ; lt9 are linearly dependent, there are real numbers�1; : : : ; �9, not all zero, such that

9XjD1

�jltj. p/ D9X

jD1�jp.tj/ D 0 for p 2 R2Œ x �3: (19.55)

Setting p D .1�x21/.1˙x2/ and p D .1�x1/x2.1˙x2/ in (19.55) yields 2�6C�9 D2�5C�9 D 0 and 2�3C�5 D 2�4C�6 D 0. If we interchange the role of x1 and x2,we have 2�7 C �9 D 2�8 C �9 D 0 and 2�2 C �7 D 0: For p D .1C x1/x2.1C x2/we get 2�1 C �5 D 0. Therefore, putting �9 D 1, we obtain on R2Œ x �3,

lt9 C1

4

�lt1 C lt2 C lt3 C lt4

� � 12

�lt5 C lt6 C lt7 C lt8

� D 0: (19.56)

(We could have avoided Chasles’ theorem by verifying this identity on R2Œ x �3 by adirect computation.) Let p 2 R2Œ x �3. By (19.56),

p.t9/2 D

�� 14

4XjD1

p.tj/C 1

2

8XjD5

p.tj/

�2� 5

4

8XjD1

:p.tj/2: (19.57)

Then Lc is a positive functional, since 0 < c � 45

and hence

Lc. p2/ D

8XjD1

p.tj/2 � cp.t9/

2 � �1 � 5c=4�8X

jD1p.tj/

2 � 0:


Since t1; � � � ; t8 are zeros of OR and OR.t9/ D 1, we have Lc. OR/ D �c < 0:

Thus, Lc is not a truncated moment functional, since R 2 Pos.R3/ and henceOR 2 Pos.R2/: ut

19.7 Exercises

1. (Covariance of the apolar scalar product)Let p; q 2 Hd;m and A 2 Md.R/. Define . p ı A/.x/ D p.Ax/; x 2 Rd; wherex 2 Rd is a column vector. Prove that pıA 2 Hd;m and Œ pıA; q� D Œ p; qıAT�.Hint: It suffices to verify the latter for p D .a�/m; q D .b�/m, where a; b 2 Rd.

2. Let p 2 Hd;2n. Show that p ıU D p for all orthogonal transformations U if andonly if there is a number c 2 R such that p.x/ D ckxk2n.

3. Define Da WD a1@1C� � �Cad@d for a D .a1; : : : ; ad/ 2 Rd. Let y1; : : : ; ym 2 Rd.Show that 1

mŠDy1 � � �Dymf D Œ f ; .y1�/ � � � .ym�/� for f 2 Hd;m:

4. Let p 2 Hd;m;m 2 N; and x 2 Rd: Suppose that @[email protected]/ D 0 for j D 1; : : : ; d:

Use Euler’s identity to show that p.x/ D 0.5. Show that the polynomial f .x1; x2/ D x21x

22 2 H2;4 is not in Q2;4.

6. Let f".x1; x2/ D x21x22 C ".x21 C x22/

2 2 H2;4 for " > 0. Note that f" is positive onR2nf0g. Show that f" is not in Q2;4 if 0 < " < 1

4.

Hint: Assume the contrary. Write f" D P5jD1.ajx1 C bjx2/4 by Carathéodory’s

theorem. Compare first the coefficients of x41, x42, and finally of x21x

22.

7. Let d D 2 or n D 1. Let p be a homogeneous polynomial of Hd;2n such thatp � 0 on Rd: Show that p is a sum of squares of elements of Hd;n.

8. Find polynomials p 2 H3;6 and q 2 H4;4 which are not sums of squares ofelements of H3;3 and H4;2, respectively.Hint: Use homogenized Motzkin and Choi–Lam polynomials, (13.6)and (13.41).

9. Suppose that d � 3; n � 3 or d � 4; n � 2. Show thatP

H2d;n ¤

Pos.Hd;2n;Rd/:

10. Find a polynomial p 2 H2;2 such that p2 does not span an extreme ray ofP2

2;4.11. Let t1; : : : ; t8 be the points of R2 given by (19.53) and (19.54) and let L DP8

jD1mjltj , where mj � 0 for j D 1; : : : ; 8:a. Show the truncated moment functional L on R2Œ x �6 is determinate.b. Let t9 2 R2 be such that t9 ¤ tj for j D 1; : : : ; 8. Show that � D ıt9 CP8

jD1mjıtj is an ordered maximal mass measure.

12. Let R be the Robinson polynomial. Find a linear transformation A of R3 suchthat p 2 H3;6 defined by p.x/ WD R.Ax/ has exactly 10 zeros in R2.

19.8 Notes 497

13. (Hessians at zeros for sums of squares)Let f1; : : : ; fk 2 H2;n and f D f 21 C � � � C f 2k . Suppose that t 2 Z. f /. Show that

dXi;jD1

�i�j@2f

@[email protected]/ D 2

kXlD1

� dXi

�i@[email protected]/

�2� 2

kXlD1.� � rfl.t//2

for � D .�1; : : : ; �d/T 2 Rd, where r D . @@x1; : : : ; @

@xd/T .

Hint: First verify that @2f 2l@xi@xj

.t/ D 2 @[email protected]/ @fl

@xj.t/:

14. Let t1; : : : ; t9 be pairwise distinct points of R2 and let f ; g 2 R2Œ x �3 be twocubics satisfying Z. f / \ Z.g/ D ft1; : : : ; t9g. Show that no four of the pointst1; : : : ; t9 are collinear and no seven are on a quadratic.

15. ([Re1]) Suppose that a > 0; a ¤ p2: Let L be the moment functionalof the measure � D P8

jD1 ıtj on R2Œ x �6, where the eight points tj are.˙1;˙1/; .˙a; 0/; .0;˙a/:a. Show that f D x1.x21�a2 C .a2�1/x22/; g D x2.x22�a2 C .a2�1/x21/ form a

basis of the vector space NL:

b. What happens to NL in the case a D p2?c. Show that Z. f / \Z.g/ D ft1; : : : ; t8; t9g, where t9 WD .0; 0/.d. Let L0 WD L C lt9 and L00 D L0 C ıt10 , where t10 D .2; 1 � 3.1 � a2/�1/ if

a ¤ 1 or t10 D .3; 3/: Determine NL0 , NL00 , NC.L0/, NC.L00/, VC.L0/, andVC.L00/:

19.8 Notes

The apolar scalar product goes back to the 19th century, see [Re1] for somehistorical discussion and [Veg], [Re3]. Representing homogeneous polynomials ofdegree m as finite sums of m-th powers of linear forms is called Waring’s problem,see [El1, El2]. Hilbert’s Theorem 19.15 was proved in [H2]. It is developed in [Nat]where explicit formulas for h4;2n; n D 2; 3; 4; 5; are given.

Robinson’s polynomial was discovered by R.M. Robinson in [Rb]. Its extremal-ity (Corollary 19.22) was shown in [CL]. Proposition 19.25 is taken from [Sm10].

Theorem 19.32 is due to M.D. Choi, T.Y. Lam and B.Reznick [CLR]. Recentresults on Carathéodory numbers can be found in [RiS] and [DSm2].

The construction method of a positive functional that is not a moment functionalin Proposition 19.39 (with a more complicated polynomial) was invented by theauthor [Sm2]. It was elaborated in [Bl1]. Formulas (19.56) and (19.57) appeared in[BW, Exercise 1.6.28].

Appendix

A.1 Measure Theory

For all notions and results on measure theory we refer to [Ba, BCRl], and [Ru2].Throughout this appendix, X is a locally compact Hausdorff space.We denote by Cc.X / the continuous functions on X with compact support, by

Cc.X IR/ the real-valued functions in Cc.X /, by Cc.X /C the nonnegative functionsin Cc.X /, and by C0.X / the functions f 2 C.X / which vanish at infinity, orequivalently, for which the set fx 2 X W jf .x/j � "g is compact for all " > 0.

The Borel algebraB.X / is the �-algebra generated by the open subsets of X . AB.X /-measurable function on X is called a Borel function.

Definition A.1 A Radon measure on X is a measure � W B.X / ! Œ0;C1� suchthat �.K/ <1 for each compact subset K of X and

�.M/ D sup f�.K/ W K M; K compactg for all M 2 B.X /: (A.1)

Thus, in our terminology Radon measures are always nonnegative!Condition (A.1) is the regularity of �. If the locally compact space X is �-

compact (that is, X is a countable union of compact sets), then each Radon measureis also outer regular (see e.g. [Ba, Corollary 29.7])

�.M/ D inf f�.U/ W M U; U openg for M 2 B.X /:

Closed subsets of Rd are obviously �-compact. All measures for moment problemsoccuring in this book are Radon measures on �-compact locally compact Hausdorffspaces; in most cases they are supported on closed subsets of Rd:

Let MC.X / denote the set of Radon measures on X , M1C.X / the subset ofprobability measures in MC.X /, and MbC.X / the subset of finite measures inMC.X /.

© Springer International Publishing AG 2017K. Schmüdgen, The Moment Problem, Graduate Texts in Mathematics 277,DOI 10.1007/978-3-319-64546-9

499

500 Appendix

A complex Radon measure on X is a map � W B.X / ! C which is of the form� D �1 � �2 C i.�3 � �4/, where �1; �2; �3; �4 2 MbC.X /: The set of complexRadon measures on X is denoted by M.X /:

For � 2 MC.X / we denote the �-integrable Borel functions on X by L1.X ; �/,that is, a Borel function f on X is in L1.X ; �/ if and only if

R jf .x/j d� <1.We say that a measure � 2 MC.X / is supported on a set M 2 B.X / if

�.XnM/ D 0. The support of � 2 MC.X /, denoted by supp �, is the smallestclosed subset M of X such that X is supported on M. By this definition, x0 2 X isin supp� if and only if �.U/ > 0 for each open subset U of X containing x0.

Theorem A.2 (Lebesgue’s Dominated Convergence Theorem) Suppose that� 2 MC.X /: Let fn; n 2 N, and f be complex Borel functions on X and letg W X ! Œ0;C1� be a �-integrable function. Suppose that

limn!1 fn.x/ D f .x/ and jfn.x/j � g.x/; n 2 N; �-a.e. on X : (A.2)

Then the functions f and fn are �-integrable,

limn!1

Z˝

jfn � f j d� D 0; and limn!1

Z˝

fn d� DZ˝

f d�:

Proof [Ru2, Theorem 1.34]. utTheorem A.3 (Radon–Nikodym Theorem) Let �; � 2 MC.X /. Suppose that �is �-finite (that is, X is a countable union of sets which have finite �-measure). If �is absolutely continuous with respect to � (that is, each �-null set is also a �-nullset), then there exists a nonnegative function h 2 L1.X ; �/ such that d� D h.x/d�.

Proof [Ba, Theorem 17.10] or [Ru2, Theorem 6.9]. utTheorem A.4 (Riesz’ Representation Theorem) For each linear functional L onCc.X IR/ such that L. f / � 0 for all f 2 Cc.X /C there exists a unique (positive)Radon measure � on X such that

F. f / DZXf d�; f 2 Cc.X IR/:

Proof [Ru2, Theorem 2.14], [Ba, Theorems 29.1 and 29.3], or [BCRl, Chapter 2,Corollary 2.3]. ut

Now we develop some basics on the vague convergence of measures, see e.g.[Ba, � 30 and 31] or [BCRl, Chapter 2, � 3 and 4] for more details.

The vague topology on MC.X / is the coarsest topology for which all mappings� 7! R

fd�, where f 2 Cc.X /; are continuous. A net .�i/i2I fromMC.X / convergesvaguely to � 2 MC.X / if and only if for all f 2 Cc.X IR/ we have

limi

Rfd�i D

Rfd�: (A.3)

Appendix 501

Proposition A.5 A net .�i/i2I of measures �i 2 MC.X / converges vaguely to� 2 MC.X / if and only if

lim supi

�i.K/ � �.K/ and lim infi

�i.A/ � �.A/

for every compact subset K and every relatively compact open subset A of X .

Proof [Ba, Theorem 30.2], see also [BCRl, Chapter 2, Exercise 4.11]. utTheorem A.6 A subset M of MC.X / is relatively compact in the vague topology(that is, its closure is vaguely compact) if and only if it is vaguely bounded, that is,

sup�2M

ˇ Rf d�

ˇ<1 for all f 2 Cc.X IR/:

For any a > 0 the set f� 2 MC.X / W �.X / � ag is vaguely compact.Proof [Ba, Theorem 31.2 and Corollary 31.3] or [BCRl, Chapter 2, Theorem 4.5and Proposition 4.6]. utProposition A.7 If Cc.X IR/ is separable with respect to the supremum norm, thenthe vague topology on MC.X / is metrizable.

Proof [Ba, Lemma 31.4 and Theorem 31.5]. utA function h W X ! Œ�1;C1� is called upper semicontinuous (resp. lowersemicontinuous) if for each number a 2 R the set fx 2 X W f .x/ < ag (resp.fx 2 X W f .x/ > ag) is open in X , or equivalently, if for each convergent net .xi/i2Iin X we have

lim supi f .xi/ � f .limixi/ .resp: f .lim

ixi/ � lim infi f .xi//:

The following result is called the portmanteau theorem.

Proposition A.8 Let .�i/i2I be a net of measures �i 2 MbC.X / and � 2 MbC.X /.Then the following five statements are equivalent:

(i) lim supi �i D � in the vague convergence and limi �i.X / D �.X /:(ii) lim supi �i.A/ � �.A/ for all closed subsets A of X and limi �i.X / D �.X /:

(iii) lim supiRh d�i �

Rh d� for all upper semicontinuous bounded real func-

tions h on X :(iv) lim infi �i.A/ � �.A/ for all open subsets A of X and limi �i.X / D �.X /:(v) lim infi

Rh d�i �

Rh d� for all lower semicontinuous bounded real functions

h on X :

Proof By [BCRl, Chapter 2, Proposition 4.2] or [Ba, Theorem 30.8], (i) holds if andonly if the net .�i/ converges weakly to �. Using this result the assertion followsfrom [BCRl, Chapter 2, Theorem 3.1]; for (i) and (iii), see [Ba, Theorem 30.10]. ut

502 Appendix

Proposition A.9 Let .�i/i2I be a net from MbC.X / satisfying supi2I �i.X / <1: Ifthis net converges vaguely to � 2 MbC.X /, then (A.3) holds for all f 2 C0.X /.

Proof [Ba, Theorem 30.6] or [BCRl, Chapter 2, Proposition 4.4]. ut

A.2 Pick Functions and Stieltjes Transforms

Let CCDfz 2 C W Im z > 0g denote the upper half-plane.

Definition A.10 A holomorphic function f W CC ! C is a Pick function if

Im f .z/ � 0 for all z 2 CC:

We denote the set of Pick functions by P. In the literature Pick functions are alsocalled Nevanlinna functions.

Examples of Pick functions are tan z or the principal logarithm Log z.If f 2 P and f .z0/ is real for some point z0 2 CC, then f is not open and hence a

constant. That is, all nonconstant Pick functions map CC into CC.Each Pick function f can be extended to a holomorphic function on CnR by

setting f .z/ WD f .z/ for z 2 CC.

Theorem A.11 (Canonical Integral Representation of Pick Functions) For eachPick function f there exist numbers a; b 2 R, b � 0; and a finite Radon measure �on the real line such that

f .z/ D aC bzCZR

1C zt

t�z d�.t/; z 2 C=R; (A.4)

where the numbers a; b and the measure � are uniquely determined by f .Conversely, any function f of this form is a Pick function.

Proof [AG, Nr. 69, Theorem 2] or [Dn, p. 20, Theorem 1]. utOften the canonical representation (A.4) is written in the form

f .z/ D aC bzCZR

�1

t�z �t

1C t2

�d�.t/; z 2 C=R; (A.5)

where � is a Radon measure on R such thatR.1 C t2/�1d�.t/ < 1. The two

canonical representations (A.4) and (A.5) are related by the formula

d�.t/ D .1C t2/�1d�.t/:

Moreover, a D Re. f .i// and b D limy!1 f .iy/iy . It should be emphasized that the

function t.1C t2/�1 in (A.5) is not necessarily �-integrable.

Appendix 503

Definition A.12 The Stieltjes transform of a complex Radon measure� 2 M.R/ is

I�.z/ DZR

1

t � zd�.t/; z 2 CnR: (A.6)

Stieltjes transforms are also called Cauchy transforms or Borel transforms.

Theorem A.13 (Stieltjes–Perron Inversion Formula) Each complex Radon mea-sure � 2 M.R/ is uniquely determined by the values of its Stieltjes transform I�.z/on CnR. In fact, for a; b 2 R; a < b, we have

�..a; b//C 12�.fag/C 1

2�.fbg/ D lim

"!C012� i

Z b

aŒI�.tCi"/�I�.t�i"/� dt;

(A.7)

�.fag/ D lim"!C0 � i" I�.aCi"/: (A.8)

In particular, if I�.z/ D 0 for all z 2 CnR, then � D 0.Proof By definition � 2 M.R/ is a linear combination of measures from MbC.R/.Hence one can assume without loss of generality that � 2 MbC.R/. In this caseproofs are given (for instance) in [AG, Nr. 69] and [Wei, Appendix B]. ut

Note that there exist complex Radon measures � ¤ 0 for which I� D 0 on CC.Now suppose that � 2 MC.R/. Then I�.z/ D I�.z/ for all z 2 C nR. Hence the

measure � is uniquely determined by the values of its Stieltjes transform I�.z/ onCC and the formulas (A.7) and (A.8) can be written as

�..a; b//C 12�.fag/C 1

2�.fbg/ D lim

"!C01�

Z b

aIm I�.tCi"/ dt; (A.9)

�.fag/ D lim"!C0 " Im I�.aCi"/: (A.10)

The Stieltjes transform I� of � 2 MC.R/ is a Pick function, since

Im I�.z/ D Im zZR

1

jt � zj2 d�.t/; z 2 CnR:

The next result characterizes these Stieltjes transforms among Pick functions.

Theorem A.14 A Pick function f is the Stieltjes transform I� of a measure � 2MC.R/ if and only if

sup f jyf .iy/j W y 2 R; y � 1g <1: (A.11)

Proof [AG, Nr. 69, Theorem 3]. ut

504 Appendix

Proposition A.15 Let K be a closed subset of R and � 2 MC.R/. The Stieltjestransform I�.z/ has a holomorphic extension to CnK if and only if supp � K.

Proof [Dn, Lemma 2, p. 26]. utLet us sketch the proof of Proposition A.15:If supp� K, then (A.6) with z 2 CnK defines a holomorphic extension of I� to

CnK. Conversely, suppose that I� has a holomorphic extension, say f , to CnK. Thenlim"!C0 I�.t ˙ i"/ D f .t/ and hence lim

"!C0 Im I�.tCi"/ D 0 for t 2 RnK. Therefore,

by (A.9) and (A.10), we have supp � K:

A.3 Positive Semidefinite and Positive Definite Matrices

Let K be either the real field R or the complex field C. If not stated otherwise, h�; �iand k � k denote the standard scalar product and the Euclidean norm of Kn.

Let Mn;m.K/ be the .n;m/-matrices over K and Mn.K/ WD Mn;n.K/. For a matrixA D .ajk/ 2 Mn;m.K/we denote by A the matrix A D .ajk/ and by AT the transposedmatrix. A matrix A D .ajk/nj;kD1 is called Hermitian if ajk D akj for j; k D 1; : : : ; n.

Proofs of the following Propositions A.16 and A.18–A.20 and the correspondingnotions can be found in standard texts on matrices such as [Gn, Zh].

Proposition A.16 (Spectral theorem for Hermitian matrices) Let A 2 Mn.K/ beHermitian. There exist an orthonormal basis u1; : : : ; un of Kn and reals �1; : : : ; �nsuch that Auj D �juj, j D 1; : : : ; n. If D denotes the diagonal matrix with diagonalentries �1; : : : ; �n and U the matrix with columns u1; : : : ; un, then

A DnX

jD1�juj.uj/

T D UDU�1: (A.12)

Given a Hermitian matrix A D .ajk/nj;kD1, the expression

QA.�/ D hA�; �i DnX

j;kD1ajk�j�k for � D .�1; : : : ; �n/T 2 Kn

is called the Hermitian form associated with A or briefly a Hermitian form.

Definition A.17 A Hermitian matrix A and the Hermitian form QA are called

� positive semidefinite if QA.�/ � 0 for all � 2 Kn,� positive definite if QA.�/ > 0 for all � 2 Kn; � ¤ 0:We write A � 0 if A is positive semidefinite and A � 0 if A is positive definite.

Given integers ij, j D 1; : : : ; k � n; such that 1 � i1 < i2 < � � � < ik � nlet D.i1; : : : ; ik/ denote the determinant of the k � k-submatrix of A formed by the

Appendix 505

ij-th rows and columns of A: Then D.i1; : : : ; ik/ is called a principal minor of A andD.1; : : : ; k/ is a main principal minor of A:

Proposition A.18 For a Hermitian matrix A 2 Mn.K/ the following are equiva-lent:

(i) A is positive semidefinite.(ii) All principal minors D.i1; : : : ; ik/ are nonnegative.

(iii) All eigenvalues of A are nonnegative.(iv) There exist vectors v1; : : : ; vn 2 Kn such that A DPn

jD1 vj.vj/T .

Proposition A.19 For a Hermitian matrix A 2 Mn.K/ the following are equiva-lent:

(i) A is positive definite.(ii) All main principal minors D.1; : : : ; k/, k D 1; : : : ; n, are positive.

(iii) All eigenvalues of A are positive.(iv) There are linearly independent vectors v1; : : : ; vn 2 Kn such that A DPk

jD1 vj.vj/T .(v) A is positive semidefinite and detA ¤ 0:

(vi) A is positive semidefinite and rankA D n.

For each positive semidefinite matrix A there is a unique positive semidefinitematrix B, denoted by A1=2, such that B2 D A. If A is of the form (A.12) and D1=2

denotes the diagonal matrix with entries �1=21 ; : : : ; �1=2n , then A1=2 D UD1=2U�1.

The trace Tr A of a matrix A 2 Mn.K/ is the sum of all diagonal entries of A.

Proposition A.20 For A;B 2 Mn.K/ we have:

(i) TrAB D TrBA.(ii) TrA � 0 if A � 0. In particular, TrA

TA � 0:

(iii) A � 0 and TrA D 0 imply that A D 0. In particular, TrATA D 0 implies

A D 0.Let Symn denote the real vector space of symmetric matrices in Mn.R/. There is

a scalar product h�; �i on Symn defined by

hA;Bi WD TrAB DnX

i;jD1aijbij for A D .aij/; B D .bij/ 2 Symn:

For matrices from Symn we have the following basic properties.

Proposition A.21

(i) If A � 0 and B � 0, then hA;Bi D TrAB � 0:(ii) If A � 0; B � 0, and hA;Bi D 0, then AB D 0.

(iii) If hA;Bi � 0 for all B � 0, then A � 0.

506 Appendix

Proof

(i) As noted above, each positive semidefinite matrix C 2 Symn has a uniquepositive semidefinite square root C1=2. Then we derive

TrAB D TrA1=2A1=2B1=2B1=2 D TrB1=2A1=2A1=2B1=2 D Tr .A1=2B1=2/T .A1=2B1=2/:

Since .A1=2B1=2/T.A1=2B1=2/ � 0, this yields TrAB � 0.(ii) By the preceding equality, we have 0 D hA;Bi D Tr .A1=2B1=2/T.A1=2B1=2/.

Therefore, .A1=2B1=2/T.A1=2B1=2/ D 0 and hence A1=2B1=2 D 0. Thus weobtain AB D A1=2.A1=2B1=2/B1=2 D 0:

(iii) Assume to the contrary that A is not positive semidefinite. Then, by Proposi-tion A.18, A admits an eigenvalue � < 0, so Au D �u with u 2 Rn, u ¤ 0:

Then B WD uuT � 0 and TrAB D Tr .�uuT/ D �Tr uuT . Since uuT � 0

and uuT ¤ 0, Tr uuT > 0 and hence TrAB < 0, which contradicts theassumption. ut

A.4 Positive Semidefinite Block Matrices and Flat Extensions

Suppose that A D AT 2 Mn.R/, B 2 Mn;m.R/, C D CT 2 Mm.R/, where n;m 2 N.We consider the symmetric block matrix X 2 MnCm.R/ defined by

X D�A BBT C

�: (A.13)

Definition A.22 The block matrix X is called a flat extension of A if

rankX D rankA:

Proposition A.23 The block matrix X is a flat extension of A if and only if thereexists a matrix U 2 Mn;m.R/ such that B D AU and C D UTAU.

Proof The proof is based essentially on the well-known fact that the rank of amatrix is the maximal number of linearly independent columns. First suppose thatrankX D rankA. Then each column of

�BC

�is in the span of columns of

�ABT

�. Hence�

BC

�is of the form

�AUBTU

�, so that B D AU and C D BTU D UTAU.

Conversely, if X is of the prescribed form, one easily verifies that the dimensionsof the ranges of X and A are equal; hence rankX D rankA. ut

The following description of positive semidefinite block matrices is due to J.L.Shmuljan [Sh].

Theorem A.24 Suppose that A � 0:(i) X � 0 if and only if B D AU for some matrix U 2 Mn;m.R/ and C � UTAU.

Appendix 507

(ii) Suppose that B D AU with U 2 Mn;m.R/: Then X is a flat extension of A if andonly if C D UTAU.

(iii) If X is an arbitrary flat extension of A, then X � 0.Proof

(i) First suppose that B D AU with U 2 Mn;m.R/. Then we have the identity

X D�

A AUUTA C

�D (A.14)

�A1=2 A1=2U0 0

�T �A1=2 A1=2U0 0

�C�0 0

0 C �UTAU

�: (A.15)

If C � UTAU, both summands in (A.15) are positive semidefinite, so thatX � 0.

Conversely, assume that X � 0: We show that B can be written as B D AUwith U 2 Mn;m.R/. Since X � 0, it has a positive square root. If X1 and X2denote the rows of X1=2 we write

X1=2 D�X1X2

�:

Comparing the square of X1=2 with X yields A D X1XT1 ;B D X1XT

2 ;C D X2XT2 :

For v 2 Rn we have hAv; vi D hX1XT1 v; vi D kXT

1 vk2: Therefore, Av D 0

implies that XT1 v D 0. Hence there exists a well-defined (!) linear operator

V1 W Rn ! RnCm such that V1Av D XT1 v for v 2 Rn. Then V1A D XT

1 :

Similarly, there is a linear operator V2 W Rm ! RnCm such that V2C D XT2 :

Then VT1 V2C W Rm ! Rn is given by a matrix U 2 Mn;m.R/ and we compute

B D X1XT2 D .V1A/TV2C D AVT

1 V2C D AU:For v 2 Rm, we have h.C � UTAU/v; vi D hX � 0v � ; � 0v �i � 0: Thus,

C � UTAU.(ii) By the assumption of (ii), B D AU. Therefore, if C D UTAU, then X is a flat

extension of A by Proposition A.23.Conversely, suppose C ¤ UTAU. Let P W RnCm ! Rn be the canonical

projection given by P.x; y/ D x, x 2 Rn; y 2 Rm. From (A.14) it follows thatP is a surjection of the range of X on the range of A. We choose v 2 Rm suchthat Cv ¤ UTAUv. Set u WD �Uv and w WD Cv � UTAUv. From (A.14) wederive X . uv / D

�0w

�. Since w ¤ 0, P is not injective. Therefore, rankX D

dim imX > dim imA D rankA, so X is not a flat extension of A.(iii) By Proposition A.23, each flat extension X of A is of the form (A.13) with

B D AU and C D UTAU. Hence X � 0 by (A.14) and (A.15). ut

508 Appendix

Proposition A.25 Suppose that A � 0. Then the matrix X given by (A.13) is a flatextension of A if and only if

RnCm D .Rn; 0/C kerX: (A.16)

If this holds and x 2 Rn, then x 2 kerA if and only if .x; 0/ 2 kerX.

Proof First suppose that X is a flat extension of A. Then, by Proposition A.23, thereis a matrixU 2 Mn;m.R/ such thatB D AU andC D UTAU. Let x 2 Rn and y 2 Rm.Then X.�Uy; y/ D 0, so that .x; y/ D .xCUy; 0/C .�Uy; y/ 2 RnCmCkerX: Thisproves (A.16).

Now assume that (A.16) holds. By reordering the canonical basis of Rn, we mayassume that the first n0 canonical basis elements of Rn are linearly independent ofkerX and that, together with a basis of kerX, they form a basis of RnCm. Then

RnCm D .Rn0

; 0/˚ kerX: (A.17)

Let m0 WD dim ker X. Clearly, n0 C m0 D nC m by (A.17). From (A.16) it followsthat n0 � n. We write X as block matrix of the form (A.13) with respect to thedecomposion (A.17). Let A0;B0; .B0/T ;C0 denote the corresponding matrix entries.The matrix of change of bases between the canonical basis of RnCm and this newbasis is of the form

�I �U00 I

�

for some matrix U0 2 Mn0;m0.R/: Then the matrix of the symmetric operatorcorresponding to X on RnCm in the new basis is

�A0 00 0

�D�

I 0

�.U0/T I

��A0 B0.B0/T C0

��I � U00 I

�

D�

A0 B0 � A0 U0.B0/T � .U0/T A0 C0�.U0/TB0 � .B0/TU0 C .U0/TA0 U0

�:

Thus, B0 D A0 U0, .B0/T D .U0/T A0; and C0�.U0/TB0 � .B0/TU0C .U0/TA0 U0 D 0:We insert the first two relations into the last and obtain C0 D .U0/TA0 U0. Hence, byProposition A.23, X is a flat extension of A0, so that rankX D rankA0: Since n0 � nas noted above, A0 corresponds to a submatrix of A in the new basis. Therefore,rankA0 � rankA � rankX: Hence rankX D rankA; so X is a flat extension of A.

We verify the last assertion. We write X as in Proposition A.23 and computeX.x; 0/ D .Ax;BTx/ D .Ax;UTAx/: Hence .x; 0/ 2 kerX if and only ifx 2 kerA. ut

Appendix 509

A.5 Locally Convex Topologies

Introductions to locally convex spaces are given in [Ru1] and [Cw].In this section, K D R or K D C and V is a K-vector space.A map p W V ! RC is called a seminorm on the vector space V if p.�x/ D

j�jp.x/ and p.xC y/ � p.x/C p.y/ for all � 2 K and x; y 2 V:Let P be a family of seminorms on V such that p.x/ D 0 for p 2 P implies x D 0.

The locally convex topology defined by P is the topology on V for which the sets

fy 2 V W p1.x � y/ � "; : : : ; pk.x � y/ � "g

where p1; : : : ; pk 2 P; k 2 N; " > 0, form a base of neighbourhoods of x 2 V .

Theorem A.26 (Separation Theorem for Convex Sets) Suppose that V is realvector space equipped with a locally convex topology. Let X and Y be nonemptydisjoint convex subsets of V.

(i) If Y is open, then there exist a linear functional L W V ! R and an a 2 R suchthat L.x/ � a < L.y/ for x 2 X and y 2 Y.

(ii) If X is compact and Y is closed, then there are a linear functional L W V ! R

and numbers a; b 2 R such that L.x/ � a < b � L.y/ for all x 2 X and y 2 Y.

If Y is a cone, then a < 0 in (i) and (ii).

Proof [Ru1, Part 1, Theorem 3.4]. utThe following algebraic version of the separation theorem is Eidelheit’s theorem.

A point x0 of a set X in a real vector space V is called an internal point if, givenv 2 V , there exists a ıv > 0 such that x0 C ıv 2 X for all ı 2 R; jıj � ıv .Theorem A.27 Let X and Y be nonempty disjoint convex sets in a real vectorspace V. Suppose that Y has at least one internal point. Then there exists a linearfunctional L on V such that L.x/ � L.y/ for x 2 X, y 2 Y. If Y is a cone, then L isY-positive.

Proof [KNa, Theorem 3.8] or [Kö], � 17, (3), p. 187. utIf the vector space V is finite-dimensional, it has a unique locally convex

topology; we call it the natural topology of V . It can be given by any norm on V .Let �f denote the finest locally convex topology on V . It is obtained by taking

the family of all seminorms on V as P. A neighbourhood base of zero is givenby the absolutely convex absorbing subsets of V . All linear functionals on V andall seminorms on V are continuous in the topology �f . The topology �f inducesthe natural topology on each finite-dimensional linear subspace. If V has countablevector space dimension, we have the following characterization of closed subsets.

Proposition A.28 Suppose that V is a K-vector space of at most countabledimension. A subset C of V is closed in the topology �f if and only C \ E is closedin V for each finite-dimensional subspace E of V.

510 Appendix

Proof The assertion follows from the Krein–Smuljan theorem [KNa, Theorem 22.6]applied to the dual locally convex space of V . We include a direct “elementary”proof following [Ms1, Section 3.6].

If C is �f -closed, then C \ E is closed in the induced topology. Since the finite-dimensional space E has a unique locally convex topology, the induced topology of�f is the natural topology of E.

We prove the converse implication. Since the assertion is trivial if V has finitedimension, we can assume that V has a countable basis, say fvn W n 2 Ng. Let Vn bethe linear span of v1; : : : ; vn. For numbers "1 > 0; : : : ; "n > 0 we abbreviate

U."1; : : : ; "n/ WD˚v D

nXjD1

�jej W j�jj � "j for j D 1; : : : ; n �:

It suffices to show that the complement M WD VnC of C is open, that is, for eachpoint v 2 M there exists a �f -neighbourhood U such that U M. Upon translationwe can assume without loss of generality that v D 0.

We prove by induction that there is a positive sequence " D ."n/n2N such that

U."1; : : : ; "n/ M \ Vn: (A.18)

Clearly, since M\V1 is open, there is an "1 > 0 such that U."1/ M\V1: Supposethat "1; : : : ; "n are constructed such that (A.18) is satisfied.

Assume to the contrary that there is no number "nC1 > 0 such that (A.18) holdsfor nC 1. Then for each k 2 N there exists an xk DPnC1

jD1 �k;jvj 2 U."1; : : : ; "n;1k /

such that xk … M. Since all “coordinates” �k;j of xk are bounded, the sequence.xk/k2N has a convergent subsequence in VnC1. Let x be the limit of such asubsequence. By xk 2 U."1; : : : ; "n;

1k /, we have j�k;jj � "j for j D 1; : : : ; n and

j�k;nC1j � 1k . Therefore, x 2 Vn and x 2 U."1; : : : ; "n/; so that x 2 M\Vn by (A.18)

and hence x … C \ Vn. But xk 2 C \ Vn, since xk … M: Since x is the limit of asubsequence of .xk/, this contradicts the assumption that C \ Vn is closed.

Thus, by induction there exists a positive sequence " satisfying (A.18). SinceU D [1

nD1U."1; : : : ; "n/ is an absolutely convex and absorbing set, U is aneighbourhood of zero in the topology �f . By (A.18) we have U M. utRemark A.29 The preceding proof shows the following result: Let .Vn/n2N be asequence of finite-dimensional subspaces of V such that Vn VnC1 for n 2 N andV D [1

nD1Vn. Then M is closed in the finest locally convex topology of V if andonly if M \ Vn is closed in Vn for all n 2 N. ı

A.6 Convex Sets and Cones

The basics of convex sets are developed in the monographs [Sn, Rf], and [Bv].Let E be a real vector space. We denoted its dual vector space by E�.

Appendix 511

Definition A.30 A subset C of E is called

� convex if �xC .1 � �/y 2 C for x; y 2 C and � 2 Œ0; 1�,� a cone if xC y 2 C and �x 2 C for x; y 2 C and � � 0:Definition A.31 For a cone C in E the dual cone of C is the set

C^ D fF 2 E� W L.x/ � 0 for x 2 Cg: (A.19)

Clearly, C^ is a cone in the real vector space E�.By the convex hull resp. conic hull of a subset X of E we mean the smallest

convex set resp. cone which contains X .From now on we assume that E is a finite-dimensional real vector space. All

topological notions refer to the unique norm topology of E.

Proposition A.32 If C is a closed cone of E, then C D C^^.

Proof This is a special case of the bipolar theorem [Cw, Theorem V.1.8]. It followsat once from separation of convex sets. Obviously, C C^^ by definition.Assume to the contrary that there exists an x 2 C^^nC. Since C is closed, byTheorem A.26(ii) there is a linear functional L on E such that L.x/ < 0 andL.y/ � 0 for y 2 C. Then L 2 C^ and hence L.x/ � 0, because x 2 C^^. Thisis a contradiction. ut

A point x 2 E is called a relatively interior point of a set C if x is in the interiorof C in the vector space spanned by C.

Proposition A.33 Suppose that C is a non-empty convex subset of E.

(i) S has relatively interior points. If E is the span of C, then C has interior points.(ii) The set C and its closure C have the same interior points.

Proof

(i) [Rf, Theorem 6.2] or [Sn, Theorem 1.1.13].(ii) [Rf, Theorem 6.3] or [Sn, Theorem 1.1.15(a)]. utProposition A.34 Suppose that C is a cone in E.

(i) If C is closed and x0 2 E is not in C, then there exists a linear functional F onE such that F.x0/ < 0 and F.x/ � 0 for x 2 C.

(ii) A point x0 2 E is a boundary point of C if and only if there exists a linearfunctional F ¤ 0 on E such that F.x0/ D 0 and F.x/ � 0 for x 2 C. Such afunctional is called a supporting functional of C at x0.

Proof

(i) [Sn, Theorem 1.3.9] or [Rf, Corollary 11.4.2].(ii) The if part is almost obvious; for the only if part we refer to [Sn, Theorems 1.3.2

and 1.3.9] or [Rf, Theorem 11.6].

512 Appendix

Clearly, (i) is a special case of Theorem A.26(ii). In [Rf] both results are statedfor convex sets; since C is a cone, the corresponding hyperplanes pass the originwhich yields the results as stated above. ut

The following result is Carathéodory’s theorem [Ca1]. Since it is used in a crucialmanner in this book, we include a proof of it.

Proposition A.35 Let d D dimE and let X be a non-empty subset of E.

(i) Each point of the conic hull of X is a nonnegative combination of at most dpoints of X.

(ii) Each point of the convex hull of X is a convex combination of at most d C 1points of X.

Proof

(i) Each point x in the conic hull of X can be represented as

x DkX

jD1�jxj with �j � 0; xj 2 X; j D 1; : : : ; k: (A.20)

Suppose that (A.20) is a corresponding representation of x with minimal k.Assume to the contrary that k > d D dimE. Then x1; : : : ; xk are linearlydependent. Hence there are reals �1; : : : ; �k such that at least one is positiveand

kXjD1

�jxj D 0: (A.21)

Upon renumbering the points we can assume �1��11 > 0 is the minimum of all

�j��1j , where �j > 0. By (A.20) and (A.21) we have

x DkX

jD1�jxj � �1��1

1

kXjD1

�jxj DkX

jD2.�j � �1��1

1 �j/xj: (A.22)

Then �j � �1��11 �j � 0 for all j. Indeed, if �j > 0, then �1��1

1 � �j��1j by

construction. If �j < 0, this is trivial. Hence (A.22) is a representation (A.20)with k � 1 summands which contradicts the minimality of k.

(ii) The assertion follows by a similar reasoning as (i) (employing the lineardependence of xj � x1, j D 2; : : : ; k) or directly from (i) by using that x isin the convex hull of X if and only if .1; x/T is in the conic hull of .1;X/T inR˚ E. ut

Appendix 513

Next we turn to extreme rays, faces, and extreme points.

Definition A.36 A subset F of the cone C is called

� a face of C if F is a subcone such that x D yC z 2 F with y; z 2 C implies thaty; z 2 F:

� an extreme ray of C if F is a face of the form F D RC � x for some x 2 C; x ¤ 0:� an exposed face of C if there exists a linear functional L 2 C^ such that

F D FL WD fc 2 C W L.c/ D 0g: (A.23)

It is easily seen that an exposed face is indeed a face. The empty set ; and thecone C are faces. A face F ¤ ;;C is called a proper face. In general, not all properfaces are exposed faces, but they are if the cone C is finitely generated.

Let L 2 C^ and let c0 be an element of the exposed face FL defined by (A.23).Then L.c/ � 0 for all c 2 C (by L 2 C^) and L.c0/ D 0 (by c0 2 FL). Therefore, ifL ¤ 0, then L is a supporting hyperplane of C at c0 and c0 is a boundary point of C.

The extreme rays are the one-dimensional faces. Restating the definition, anextreme ray of C is a subset F D RC � x, where x 2 C; x ¤ 0; such that x D x1 C x2with x1; x2 2 C implies that x1; x2 2 RC � x: In this case, we say that the vector xspans an extreme ray of C. Let Exr.C/ denote the set of elements of all extremerays of C.

A cone C is called pointed if C \ .�C/ D f0g, or equivalently, if C does notcontain a line.

Proposition A.37 Suppose C ¤ f0g is a pointed closed cone of a finite-dimensional real vector space. Then each point of C is a finite sum of elementsof Exr.C/.

Proof [Sn, Theorem 1.4.3] or [Bv, II, Corollary 8.5 and problem 3]; see also [Rf,Corollary 18.7.1]. utOften it is convenient to study cones by means of bases.

Definition A.38 A base of a cone C is a convex subset B of C such that for eachx 2 C; x ¤ 0; there exists a unique number �x > 0 satisfying �x x 2 B.

A linear functional L on E is strictly C-positive if L.c/ > 0 for all c 2 C; c ¤ 0.

Let L be a strictly C-positive linear functional. Then it is easily verified that

B.L/ WD fc 2 C W L.c/ D 1g (A.24)

is a base of C. It can be shown that each base of C is of this form. For x 2 C; x ¤ 0;the corresponding number L.x/ is given by L.x/ D ��1

x , where �x > 0 and �x x 2 B:If E D C�C, then the map L 7! B.L/ yields a one-to-one correspondence betweenbases of C and strictly C-positive linear functionals on E.

Definition A.39 A point x 2 B is called an extreme point of a convex set B ifwhenever x D �yC .1 � �/z with y; z 2 B and 0 < � < 1, then y D z D x:

514 Appendix

Suppose that B is a base of C. We record some basic facts. A point x 2 B is anextreme point of the convex set B if and only if x spans an extreme ray of C, see[Bv, II, Lemma 8.4]. Further, if B is compact, C is closed [Bv, II, Lemma 8.6].

By a classical result of H. Minkowski, each compact convex set B in the finite-dimensional real vector space E is the convex hull of its extreme points.

Finally, we turn to conic optimization. Let L1 and L0 be linear functionals on E.A linear conic optimization problem is given by

c WD inf fL1. f / W f 2 C;L0. f / D 1g (A.25)

and the corresponding dual problem to (A.25) is

c� WD sup f� 2 R W .L1 � �L0/ 2 C^g: (A.26)

Lemma A.40 Suppose that L1;L0 2 C^ and there exists an element f0 2 C suchthat L0. f0/ > 0. Then c D c�:

Proof The set V WD ff 2 C W L0. f / D 1g is not empty, since L0. f0/�1f0 2 V . Hencec is defined. Let f 2 C. From L1.g/ � c for g 2 V and L0. f / � 0 (by L0 2 C^) weconclude that L1. f / � cL0. f /. Therefore, .L1 � cL0/ 2 C^ and hence c � c�.

Since L1 2 C^, .L1�0�L0/ 2 C^, so the corresponding set in (A.26) is not empty.Fix � � c� such that .L1��L0/ 2 C^. Then, for f 2 C, we have L1. f /��L0. f / � 0,so that L1.g/ � � for g 2 V . Taking the infimum over g 2 V and then the supremumover such � we get c � c�. ut

A.7 Symmetric and Self-Adjoint Operators on Hilbert Space

All operator-theoretic facts in this appendix can be found in most books on Hilbertspace operators such as [RS2], [AG], and [Sm9].

Suppose that H is a complex Hilbert space with scalar product h�; �i.By an operator on H we mean a linear mapping T of a linear subspace D.T/,

called the domain of T, into H. The kernel of T is N .T/ D f' 2 D.T/ W T' D 0g.The bounded operators defined on H are denoted by B.H/.

An operator T is called symmetric if hT'; i D h';T i for '; 2 D.T/:If T1 and T2 are linear operators on H, we say that T2 is an extension of T1 and

write T1 T2 if D.T1/ D.T2/ and T1' D T2' for ' 2 D.T1/.An operator T is called closed if for each sequence .'n/ from D.T/ such that

'n ! ' and T'n ! in H then ' 2 D.T/; D T'. If T has a closed extension,it has a smallest closed extension, called the closure of T and denoted by T.

For a closed operator T the resolvent set .T/ is the set of numbers z 2 C forwhich the operator T � zI has a bounded inverse .T � zI/�1 that is defined on thewhole Hilbert space H. The set �.T/ WD Cn.T/ is the spectrum of T.

Appendix 515

Suppose that D.T/ is dense in H. Then the adjoint operator T� of T is defined:Its domain D.T�/ is the set of vectors 2 H for which there exists an 2 H suchthat hT'; i D h'; i for all ' 2 D.T/; in this case T� D .

The operator T is symmetric if and only T T�:A densely defined operator T is called self-adjoint if T D T� and essentially

self-adjoint if its closure T is self-adjoint, or equivalently, if T D T�.Let us turn to self-adjoint extensions of a densely defined symmetric operator T.

The deficiency indices of T is the pair .dC.T/; d�.T// of (cardinal) numbers

dC.T/ D dimN .T� � zI/ for Im z > 0; d�.T/ D dimN .T� � zI/ for Im z < 0:

These numbers are independent of the choice of z satisfying Im z > 0 resp. Im z < 0.

Proposition A.41 Let T be a densely defined symmetric operator onH. Then

dim .D.T�/=D.T// D dC.T/C d�.T/: (A.27)

The operator T has a self-adjoint extension on H if and only if it has equaldeficiency indices, that is, dC.T/ D d�.T/.

Proof [Sm9, Proposition 3.7 and Theorem 13.10]. utIf dC.T/ D d�.T/ ¤ 0, then T has “many” different self-adjoint extension on H,

see [Sm9, Theorem 13.10]. Further, each densely defined symmetric operator has aself-adjoint extension on a possibly larger Hilbert space [Sm9, Proposition 3.17].

Proposition A.42 For any densely defined symmetric operator T on H the follow-ing are equivalent:

(i) T is essentially self-adjoint.(ii) T has a unique self-adjoint extension onH.

(iii) .T � zCI/D.T/ and .T � z�I/D.T/ are dense in H for some (then for all)zC; z� 2 C such that Im zC > 0 and Im z� < 0.

(iv) dC.T/ D d�.T/ D 0.If the symmetric operator T is positive, these conditions are equivalent to(v) .T � xI/D.T/ is dense inH for one (then for all) x < 0.

Proof [Sm9, Propositions 3.8 and 3.15 and Theorem 13.10]. utA conjugation is a mapping J W H! H such that for ˛; ˇ 2 C and x; y 2 H;

J.˛xC ˇy/ D ˛J.x/C ˇJ.y/; hJx; Jyi D hy; xi; and J2x D x: (A.28)

Proposition A.43 Let T be a densely defined symmetric operator on H. If thereexists a conjugation J such that JD.T/ D.T/ and JTxDTJx for all x 2 D.T/,then T has a self-adjoint extension onH.

Proof [Sm9, Theorem 13.25]. ut

516 Appendix

Suppose that T is a self-adjoint operator. By the spectral theorem, there existsa unique projection-valued measure ET on the Borel �-algebra B.R/ such thatT D R

R�dET.�/: The support of ET is the spectrum �.T/: For each Borel function

f on the spectrum of T there exists an operator f .T/ D R f .�/ dET.�/ defined by

f .T/' DZ

f .�/ dET.�/'; where ' 2 H andZjf .�/j2dhET.�/'; 'i <1:

The assignment f 7! f .T/ is the functional calculus of T, see [Sm9, Chapter 5].

Bibliography

[AM] Agler, J. and J.E. McCarthy: Pick Interpolation and Hilbert Function Spaces, Amer.Math. Soc., Providence, R.I., 2002.

[Ak] Akhiezer, N.I.: The Classical Moment Problem, Oliver and Boyd, Edinburgh andLondon, 1965.

[AG] Akhiezer, N.I. and I.M. Glazman: Theory of Linear Operators in Hilbert Space, Ungar,New York, 1961.

[AK] Akhiezer, N.I. and M.G. Krein: Some Questions in the Theory of Moments, Kharkov,1938; Amer. Math. Soc. Providence, R. I., 1962.

[AL] Anjos, M.A. and J.B. Lasserre (Editors): Handbook on Semidefinite, Conic and Polyno-mial Optimization, Internat. Ser. Oper. Res. and Management Sci. 166, Springer-Verlag,New York, 2012.

[Bk1] Bakan, A.: Codimension of polynomial subspaces in L2.Rd ; d�/ for discrete indetermi-nate measures, Proc. Amer. Math. Soc. 130(2002), 3545–3553.

[BR] Bakan, A. and S. Ruscheweyh: Representation of measures with simultaneous polyno-mial denseness in Lp.R; �/; 1 � p < 1, Ark. Math. 43(2005), 221–249.

[BW] Bakonyi, M. and H.J. Woerdeman: Matrix Completions, Moments, and Sums of Her-mitean Squares, Princeton Univ. Press, Princeton, NJ, 2011.

[Bv] Barvinok, A.: A course in Convexity, Amer. Math. Sov., Providence, R.I., 2002.[Ba] Bauer, H.: Measure and Integration Theory, Walter de Gruyter, Berlin, 2001.[BS] Becker, E. and N. Schwartz: Zum Darstellungssatz von Kadison–Dubois, Arch. Math.

40(1983), 42–428.[BN] Ben-Tal, A. and A. Nemirovsky: Lectures on Modern Convex Optimization, SIAM,

Philadelpia, 2001.[Bz] Berezansky, Y.M.: Expansions in Eigenfunctions of Self-adjoint Operators, Amer. Math.

Soc., Providence, R.I., 1968.[Be] Berg, C.: Markov’s theorem revisited, J. Approx. Theory 78(1994), 260–275.[BC1] Berg, C. and J.P.R. Christensen: Density questions in the classical theory of moments,

Ann. Institut Fourier (Grenoble) 31(1981), 99–114.[BC2] Berg, C. and J.P.R. Christensen: Exposants critiques dans le problem des moments, C.R.

Acad. Sci. Paris 296(1983), 661–663.[BD] Berg, C. and A.J. Duran: The index of determinacy for measures and the l2-norm of

orthonormal polynomials, Trans. Amer. Math. Soc. 347(1995), 2795–2811.[BP] Berg, C. and H.L. Pedersen: On the order and the type of the entire functions associated

with an indeterminate Hamburger moment problem, Ark. Mat. 32(1994), 1–11.


517

518 Bibliography

[BS1] Berg, C. and R. Szwarc: The smallest eigenvalue of Hankel matrices, Constr. Approx.34(2011), 107–133.

[BS2] Berg, C. and R. Szwarc: On the order of indeterminate moment problems, Adv. Math.250(2014), 105–143.

[BS3] Berg, C. and R. Szwarc: A determinant characterization of moment sequences withfinitely many mass points, Linear Multilinear Algebra 3(2015), 1568–1576.

[BT] Berg, C. and M. Thill: Rotation invariant moment problems, Acta Math. 167(1991),207–227.

[BV] Berg, C. and G. Valent: The Nevanlinna parametrization for some indeterminate Stieltjesmoment problems associated with birth and death processes, Methods Appl. Anal.1(1994), 169–209.

[BCI] Berg, C., Y. Chen and M.E.H. Ismail: Small eigenvalues of large Hankel matrices: theindeterminate case, Math. Scand. 9(2002), 7–81.

[BCJ] Berg, C., J.P.R. Christensen, and C.U. Jensen: A remark on the multidimensionalmoment problem, Math. Ann. 243(1979), 163–169.

[BCRl] Berg, C., J.P.R. Christensen, and P. Ressel: Harmonic Analysis on Semigroups, Springer-Verlag, New York, 1984.

[Bn] Bernstein, S.: Sur le representations des polynomes positifs, Soobshch. Kharkov matem.ob-va 14(1915), 227–228.

[Bl] Billingsley, P.: Probability and Measures, Wiley, New York, 1995.[Bi] Bisgaard, T.M.: The two-sided complex moment problem, Ark. Mat. 27(1989), 23–28.[Bx] Bix, R.: Conics and Cubics, Springer-Verlag, New York, 1998.[Bl1] Blekherman, G.: Nonnegative polynomials and sums of squares, J. Amer. Math. Soc.

25(2012), 617–635.[Bl2] Blekherman, G.: Positive Gorenstein ideals, Proc. Amer. Math. Soc. 43(2015),69–86.[BlLs] Blekherman, G. and J.B. Lasserre: The truncated K-moment problem for closure of open

sets, J. Funct. Anal. 23(2012), 3604–3616.[BTB] Blekherman, G., R.R. Thomas, and P.A. Parillo (Editors): Semidefinite Optimization and

Convex Algebraic Geometry, MOS-SIAM Series Optimization 13, Philadelphia, 2012.[BCRo] Bochnak, J., M. Coste and M.-F. Roy: Real Algebraic Geometry, Springer-Verlag,

Berlin, 1998.[Bou] Bourbaki, N.: Integration, Hermann, Paris, 1969.[Br] Bram J.: Subnormal operators, Duke Math. J. 22(1955), 75–94.[BCa1] Buchwalther, H. and G. Cassier: Measures canoniques dans le probleme classique des

moments, Ann. Inst. Fourier (Grenoble) 34(1984), 45–52.[BCa2] Buchwalther, H. and G. Cassier: La parametrisation de Nevanlinna dans le probleme de

moments de Hamburger, Expo. Math. 2(1984), 155–178.[BGHN] Bultheel, A., P. Gonzalez-Vera, E. Hendriksen, and O. Njastad: Orthogonal Rational

Functions, Cambridge Univ. Press, Cambridge, 1999.[BSS] Burgdorf, S., C. Scheiderer, and M. Schweighofer: Pure states, nonnegative polynomials,

and sums of squares, Comm. Math. Helvetici 87(2012), 113–140.[Ca1] Carathéodory, C.: Über den Variabilitätsbereich der Koeffizienten von Potenzreihen, die

gegebene Werte nicht annehmen. Math. Ann. 64(1907), 95–115.[Ca2] Carathéodory, C.: Über den Variabilitätsbereich der Fourier’schen Konstanten von

positiven harmonischen Funktionen, Rend. Circ. Mat. Palermo 32(1911), 193–217.[Cl] Carleman, T.: Les fonctions quasi-analytiques, Gauthier-Villars, Paris, 1926.[Cs] Cassier, G.: Problem de moments sur un compact de Rn et decomposition de polynomes

a plusieurs variables, J. Funct. Anal. 58(1984), 254–266.[Ch] Chandler, J.D., Jr.: Rational moment problem for compact sets, J. Approx. Theory

79(1993), 72–88.[CI] Chen, Y. and M.E.H. Ismail: Some indeterminate moment problems and Freud-like

weights, Constr. Appr. 14(1998), 439–458.[Chi1] Chihara, T.S.: An Introduction to Orthogonal Polynomials, Gordon and Breach, New

York, 1978.

Bibliography 519

[Chi2] Chihara, T.S.: On indeterminate symmetric moment problems, Math. Anal. Appl.85(1982), 331–346.

[ChiI] Chihara, T.S. and M.E.H. Ismail: Extremal measures for a system of orthogonalpolynomials, J. Constructive Appr. 9(1993), 111–119.

[Chq] Choquet, G.: Lectures on Analysis, vol. 3, Benjamin, Reading, 1969.[CL] Choi, M.D. and T.Y. Lam: Extremal positive semidefinite forms, Math. Ann. 231(1977),

1–18.[CLR] Choi, M.D., T.Y. Lam and B. Reznick: Real zeros of positive semidefinite forms, Math.

Z. 171(1987), 559–580.[Chl] Christoffel, E.B.: Über die Gaussische Quadratur und eine Verallgemeinerung derselben,

J. reine angew. Math. 55(1858), 61–82.[CMN] CimpriLc, J., M. Marshall, M., and T. Netzer: On the real multidimensional rational K-

moment problem, Trans. Amer. Math. Soc 363(2011), 5773–5788.[Cd] Coddington, E.A.: Formally normal operators having no normal extensions, Canad. J.

Math. 17(1965), 1030–1040.[Cw] Conway, J.B.: A Course in Functional Analysis, Springer-Verlag, New York, 1990.[CLO] Cox, D., J. Little and D. O’Shea: Ideals, Varieties, and Algorithms, Springer-Verlag,

New York, 1992.[CF1] Curto, R. and L. Fialkow: Recursiveness, positivity, and truncated moment problems,

Houston J. 17(1991), 603–635.[CF2] Curto, R. and L. Fialkow: Solutions of the Truncated Moment Problem for Flat Data,

Memoirs Amer. Math. Soc. 119, Providence, R. I., 1996.[CF3] Curto, R. and L. Fialkow: Flat Extensions of Positive Moment Matrices: Recursively

Generated Relations, Memoirs Amer. Math. Soc. 136, Providence, R. I., 1998.[CF4] Curto, R. and L. Fialkow: Solution of the truncated hyperbolic moment problem, Integral

Equ. Operator Theory 52(2005), 181–218.[CF5] Curto, R. and L. Fialkow: Truncated K-moment problems in several variables, J.

Operator Theory 54(2005), 189–226.[CF5] Curto, R. and L. Fialkow: An analogue of the Riesz-Haviland theorem for the truncated

moment problem, J. Funct. Anal. 255(2008), 2709–2731.[CFM] Curto, R., L. Fialkow and H.M. Möller: The extremal truncated moment problem, Int.

Equ. Operator Theory 60(2008), 177–200.[DJ] De Jeu, M.: Determinate multidimensional moment measures, the extended Carleman

condition and quasi-analytic weights, Ann. Prob. 31(2007), 1205–1227.[DM] Derkach, V. A. and M.M. Malamud: The extension theory of hermitian operators and

the moment problem, J. Math. Sci. 73(1995), 141–242.[DS] Dette, H. and W.J. Studden: The Theory of Canonical Moments with Applications in

Statistics, Probability, and Analysis, Wiley, New York, 1997.[Dv1] Devinatz, A.: Integral representations of positive definite functions, II; Trans. Amer.

Math. Soc. 77(1955), 455–480.[Dv2] Devinatz, A.: Two parameter moment problems, Duke Math. J. 24(1957), 48–498.[DF] Diaconis, P. and D. Freeedman: The Markov moment problem and de Finetti’s theorem

I, Math. Z. 247(2004), 183–199.[DSm1] di Dio, Ph. and K. Schmüdgen: On the truncated multidimensional moment problem:

atoms, determinacy, and core variety, ArXiv:1703.01497, to appear in J. Funct. Analysis.[DSm2] di Dio, Ph. and K. Schmüdgen: On the truncated multidimensional moment problem:

Carathéodory numbers, ArXiv:1703.01494, to appear in J. Math. Anal. Appl.[Dn] Donoghue, W.F. Jr.: Monotone Matrix Functions and Analytic Continuation, Springer-

Verlag, Berlin, 1974.[Do] Douglas, R.: Extremal measures and subspace density, Michigan Math. J. 11(1964),

243–246.[DX] Dunkl, C.F. and Y. Xu: Orthognal Polynomials of Several Variables, Cambridge Univ.

Press, Cambridge, 2001.

520 Bibliography

[DK] Dym, H. and V. Katznelson: Contributions of Issai Schur to analysis, in Studies inmemory of Issai Schur, Progress Math. 20, Birkhäuser, Boston, 2003, pp. xci–clxxxviii.

[EGH] Eisenbud, D., M. Green and J. Harris: Cayley–Bacharach theorems and conjectures,Bull. Amer. Math. Soc. 33(1996), 295–324.

[El1] Ellison, W.J.: A “Waring problem” for homogeneous forms, Cambridge Phil. Soc.65(1969), 663–672.

[El2] Ellison, W.J.: Waring’s problem, Amer. Math. Monthly 78(1971), 10–36.[Es] Eskin, G.I.: A sufficient condition for the solvability of a multidimensional problem of

moments. Dokl. Akad. Nauk SSSR 133(1960), 540–543.[Fv] Favard, J.: Sur le polynomes de Tchebycheff, C. R. Acad. Sci. Paris 200(1935),

2052–2055.[Fj] Fejér, L.: Über trigonometrische Polynome, J. reine angew. Math. 146(1916), 53–82.[F1] Fialkow, L.: The truncated K-moment problem: a survey, Preprint, 2015.[F2] Fialkow, L.: The core variety of a multisequence in the truncated moment problem,

Preprint, 2015.[FN1] Fialkow, L. and J. Nie.: Positivity of Riesz functionals and solutions of quadratic and

quartic moment problems, J. Funct. Analysis 258(2010), 328–356.[FN2] Fialkow, L. and J. Nie: The truncated moment problem via homogenization and flat

extensions, J. Funct. Analysis 23(2012), 1682–1700.[Fi] Fischer, E.: Über das Carathéodory’sche Problem, Potenzreihen mit positivem reellen

Teil betreffend, Rend. Circ. Mat. Palermo 32(1911), 240–256.[Fo] Foias, C.: Décomposition en opérateurs et vectors propres. I. Rev. Roumaine Math. Pures

Appl. 7(1962), 241–282.[Fr1] Friedrich, J.: A note on the two dimensional moment problem, Math. Nachr. 85(1984),

285–286.[Fr2] Friedrich, J.: Operator moment problems, Math. Nachr. 151(1991), 273–293.[FK] Fritzsche, B. and B. Kirstein (Editors): Ausgewählte Arbeiten zu den Ursprüngen der

Schur-Analysis, Teubner-Verlag, Stuttgart-Leipzig, 1991.[Fu] Fuglede, B.: The multidimensional moment problem, Expo. Math. 1(1983), 47–65.[Fn] Fulton, W.: Algebraic curves: an introduction to algebraic geometry, Addison-Wesley,

1998.[Gb] Gabardo, J.-P.: A maximum entropy approach to the classical moment problem, J. Funct.

Analysis 106(1992), 80–94.[Gn] Gantmacher, F.R.: Matrizenrechnung, DVW, Berlin, 1986.[Gr] Garnett, J.B.: Bounded Analytic Functions, Springer-Verlag, New York, 2006.[Ge] Gebhardt, R.: Das eindimensionale Momentenproblem–Lösungen endlicher Ordnung

und das Momentenproblem auf Œ0;1/, Diplomarbeit, Universität Leipzig, 2010.[Gs] Geronimus, Ya. L.: On polynomials orthogonal on the circle, on trigonometric moment

problem and on allied Carathéodory and Schur functions (Russian), Mat. Sb. 15(1944),99–130.

[GLPR] Gravin, N., J. Lasserre, D.V. Pasechnik, and D. Robins: The inverse moment problemfor convex polytopes, Discrete Comput. Geom. 48(2012),596–621.

[GHPP] Gustafsson, B., C. He, M. Peyman, and M. Putinar: Reconstruction planar domains fromtheir moments, Inverse Problems 1(2000), 1053–1070.

[Hm] Hamburger, H.L.: Über eine Erweiterung des Stieltjesschen Momentenproblems, Math.Ann. 81(1920), 235–319, and 82(1920), 120–164, 168–187.

[Hn] Handelman, D.: Representing polynomials by positive linear functions on compactconvex polyhedra, Pacific J. Math. 132(1988), 35–62.

[Hs] Hausdorff, F.: Summationsmethoden und Momentenfolgen, Math. Z. 9(1921), 74–109and 280–299.

[Ha] Haviland, E. K.: On the momentum problem for distribution functions in more than onedimension, I. Amer. J. Math. 57(1935), 562–582, and II. 58(1936), 164–168.

[He] Heine, E.: Handbuch der Kugelfunktionen. Theorie und Anwendungen, Vols. 1 and 2, G.Reimer, Berlin, 1878 and 1881.

Bibliography 521

[Hz] Herglotz, G.: Über Potenzreihen mit positivem reellen Teil im Einheitskreis, Ber. Ver.Sächs. Ges. d. Wiss. Leipzig 63(1911), 501–511.

[H1] Hilbert, G.: Über die Darstellung definiter Formen als Summe von Formenquadraten,Math. Ann. 32(1888), 342–350.

[H2] Hilbert, G.: Beweis für die Darstellbarkeit der ganzen Zahlen durch eine feste Zahl n-terPotenzen, Math. Ann. 7(1909), 281–300.

[HS] Hildebrandt, T.H. and I.J. Schoenberg: On linear functional operators and the momentproblem for a finite interval in one or several dimensions, Ann. Math 34(1933), 317–328.

[Hr] Hörmander, L.: The Analysis of Linear Partial Differential Operators I., Springer-Verlag,Berlin, 1973.

[Hu] Hurwitz, A.: Über die Komposition der quadratischen Formen von beliebig vielenVariablen, Nachr. Königl. Ges. Wiss. Göttingen, 1898.

[If] Infusino, M.: Quasi-analyticity and determinacy of the full moment problem from finiteto infinite dimensions, In: Stochastic and Infinite Dimensional Analysis, Trends in Math.,Birkhäuser, 2016.

[Ii] Isii, K.: The extrema of probability determined by generalized moments (1): boundedrandom variables, Annales Inst. Math. Statist. 12(1960), 119–133.

[Is] Ismail, M.E.H.: Classical and Quantum Orthogonal Polynomials in One Variable,Cambridge Univ. Press, Cambridge, 2004.

[IM] Ismail, M.E.H. and D.R. Masson: q-Hermite polynomials, biorthogonal rational func-tions and q-beta integrals, Trans. Amer. Math Soc. 346(1994), 63–116.

[Jac] Jacobi, K.G.J.: Über Gauss neue Methode, die Werthe der Integrale näherungsweise zufinden, J. reine angew. Math. 1(1826), 301–308.

[Jc] Jacobi, T.: A representation theorem for certain partially ordered commutative rings,Math. Z. 23(2001), 259–273.

[JP] Jacobi, J. and A. Prestel: Distinguished representations of strictly positive polynomials,J. reine angew. Math. 532(2001), 223–235.

[JT] Jones, W.B. and W.J. Thron: Continued Fractions. Analytic Theory and Applications,Addison-Wesley, Reading, Mass., 1980.

[Kd] Kadison, R.V.: A Representation Theory for Commutative Topological Algebras,Memoirs Amer. Math. Soc. 7, Providence, R.I., 1951.

[Ka] Karlin, S.: Representation theorems for positive functions, J. Math. Mech. 12(1960),599–618.

[KSh] Karlin, M. and L.S. Shapley: Geometry of Moment Spaces, Memoirs Amer. Math. Soc.12, Providence, R. I., 1953.

[KSt] Karlin, M. and W. Studden: Tchebycheff Systems: with Applications in Analysis andStatistics, Interscience, New York, 1966.

[KNa] Kelley, J.E. and I. Namioka: Linear Topological Spaces, Springer-Verlag, Berlin, 1976.[Kp1] Kemperman, J.H.B.: The general moment problem, Ann. Math. Stat. 39(1968), 93–122.[Kp2] Kemperman, J.H.B.: Geometry of the moment problem, in: Landau, H.J. (Editor)

Moments in Mathematics, pp. 110–124, Amer. Math. Soc., Providence, R.I., 1987.[Kv] Khrushchev, S.V.: Schur’s algorithm, orthogonal polynomials and convergence of Wall’s

continued fractions in L2.T/, J. Approx. Theory 108(2001), 161–248.[Ki] Kilpi, Y.: Über das komplexe Momentenproblem, Ann. Acad. Sci. Fenn A Math.

236(1957), 1–31.[KlM] Klebanov, L.B. and S.T. Mkrtchayn: Estimates of the closeness of distributions in terms

of identical moments, in: Problems of stability of stochastic models, Proc. Fourth All-Union Sem., Moscow, 1980, 64–72; Translations: J. Soviet Math. 32(1986), 54–60.

[Ks] Koosis, P.: The Logarithmic Integral I, Cambridge Univ. Press, Cambridge, 1988.[KMP] Korenblum, B., A. Mascuilli and J. Panariell: A generalization of Carleman’s uniqueness

theorem and a discrete Phragmen–Lindelöf theorem, Proc. Amer. Math. Soc. 12(1998),2025–2032.

[Kö] Köthe, G.: Topological Vector Spaces, Springer-Verlag, Berlin, 1983.

522 Bibliography

[Kr1] Krein, M.G.: On an extrapolation problem due to Kolmogorov, Doklady Akad. NaukSSSR 46(1945), 306–309.

[Kr2] Krein, M.G.: The ideas of P.L. Cebysev and A.A. Markov in the theory of limiting valuesof integrals and their further developments, Uspehi Mat. Nauk 6(1951), 2–120; Amer.Math. Transl., Amer. Math. Soc., RI, 12(1951), 1–122.

[Kr3] Krein, M.G.: The description of all solutions of the truncated power moment problemand some problems of operator theory, Mat. Issled. 2(1967), 114–132. Amer. Mat. Soc.Transl. 95(1970), 219–234.

[KN] Krein, M.G. and A.A. Nudelman: TheMarkov Moment Problem and Extremal Problems,Amer. Math. Soc., Providence, R. I, 1977.

[Kv1] Krivine, J.L.: Anneaux preordonnees, J. Analyse Math. 12(1964), 307–326.[Kv2] Krivine, J.L.: Quelques properties des preordres dans les anneaux commutatifs unitaires,

C. R. Acad. Sci. Paris 258(1964), 3417–3418.[KM] Kuhlmann, S. and M. Marshall: Positivity, sums of squares and the multi-dimensional

moment problem, Trans. Amer. Math. Soc. 354(2002), 4285–4302.[KMS] Kuhlmann, S., M. Marshall, and N. Schwartz: Positivity, sums of squares and the multi-

dimensional moment problem. II, Adv. Geom. 5(2005), 583–606.[L1] Landau, H.J.: The classical moment problem: Hilbertian proofs, J. Funct. Anal.

38(1980), 255–272.[L2] Landau, H.J. (Editor): Moments in Mathematics, Proc. Symposia Appl. Math. 37, Amer.

Math. Soc., Providence, R. I, 1987.[Ls1] Lasserre, J.B.: Global optimization with polynomials and the problem of moments,

SIAM J. Optim. 11(2001), 796–817.[Ls2] Lasserre, J.B.: Moments, Positive Polynomials and Their Applications, Imperial College

London, 2010.[Ls3] Lasserre, J.B.: The K-moment problem for continuous linear functionals, Trans. Amer.

Mat. Soc. 365(2013), 2489–2504.[Ls4] Lasserre, J.B.: Borel measures with a density on a compact semi-algebraic set, Arch.

Math. (Basel) 101(2013), 361–371.[La1] Laurent, M.: Revisiting two theorems of Curto and Fialkov on moment matrices, Proc.

Amer. Math. Soc. 133(2005), 2965–2976.[La2] Laurent, M.: Sums of squares, moment matrices and optimization, In: Emerging

Applications of Algebraic Geometry, M. Putinar and S. Sullivant (Editors), Springer,New York, 2009, 157–270.

[LM] Laurent, M. and B. Mourrain: A generalized flat extension theorem for moment matrices,Archiv Math. 93(2009), 87–98.

[Lin1] Lin, G.D.: On the moment problem, Statist. Probab. Lett. 35(1997), 85–90. Erratum50(2000), 205.

[Lin2] Lin, G.D.: Recent developments on the moment problem, arXiv:1703.01027.[Lu] Lukacs, F. : Verschärfung des ersten Mittelwertsatzes der Integralrechnung für rationale

Polynome, Math. Z. 2(1918), 177–216.[Ml] Malyshev, V.A.: Cell structure of the space of real polynomials, St. Petersburg Math. J.

15(2004), 191–248.[MA] Marcellan, F. and R. Alvarez-Nodarse: On the “Favard theorem” and its extension, J.

Comput. Appl. Math. 37(2001), 213–254.[Mv1] Markov, A.A.: Deux demonstrations de la convergence de fractions continues, Acta

Math. 19(1895), 235–319.[Mv2] Markov, A.A.: Lectures on functions deviating least from zero, Mimeographed Notes, St.

Petersburg, 1906, 244–281. Reprinted in: Selected papers on continued fractions and thetheory of functions deviating least from zero, OGIZ, Moscow, 1948, 244–281. (Russian)

[Ms1] Marshall, M.: Positive Polynomials and Sums of Squares, Math. Surveys and Mono-graphs 146, Amer. Math. Soc., Providence, R.I., 2008.

[Ms2] Marshall, M.: Representations of non-negative polynomials, degree bounds and applica-tions to optimization, Canad. J. Math. 61(2009), 205–221.

Bibliography 523

[Ms3] Marshall, M.: Application of localization to the multivariate moment problem, Math.Scand. 115(2014), 269–286.

[Mt] Matzke, J.: Mehrdimensionale Momentenprobleme und Positivitätskegel, DissertationA, Universität Leipzig, 1992.

[MS] Mourrain, B. and K. Schmüdgen: Flat extensions in �-algebras, Proc. Amer. Math. Soc.144(2016), 4873–4885.

[Mo] Motzkin, T.: The arithmetic-geometric inequality, In: Inequalities, O. Sisha (Editor),Academic Press, 1967, 205–224.

[Na] Naimark, M.A.: Extremal spectral functions of a symmetric operator, Dokl. Akad. NaukSSSR 54(1946), 7–9.

[Nat] Natanson, M.B.: Additive number theory, Springer-Verlag, New York, 1996.[Nt] Netzer, T.: An elementary proof of Schmüdgen’s theorem on the moment problem of

closed semi-algebraic sets, Proc. Amer. Math. Soc. 136(2008), 529–537.[Nv1] Nevanlinna, R.: Asymptotische Entwicklungen beschränkter Funktionen und das Stielt-

jessche Momentenproblem, Ann. Acad. Sci. Fenn. A 18(1922), 1–53.[Nv2] Nevanlinna, R.: Über beschränkte analytische Funktionen, Ann. Acad. Sci. Fenn. A 32

(1929), 1–75.[Nie] Nie, J.: Optimimality conditions and finite convergence of Lasserre’s hierarchy, Math.

Program. Ser. A 14(2014), 97–121.[Nu1] Nussbaum, A.E.: Quasi-analytic vectors, Ark. Math. 6(1965), 179–191.[OW] Overton, M. and R.S. Womersley: On the sum of the k largest eigenvalues of a symmetric

matrix, SIAM J. Matrix Anal. Appl. 13(1992), 41–45.[Pa] Parillo, P.A.: Semidefinite programming relaxations for semialgebraic problems, Math.

Program. 9(2003), 293–320.[Pd1] Pedersen, H. L.: La parametrisation de Nevanlinna at le problem des moments de

Stieltjes indetermine, Expo. Math. 15(1997), 273–278.[Pd2] Pedersen, H. L.: On Krein’s theorem for indeterminacy of the classical moment problem,

J. Approx. Theory 95(1998), 90–100.[Pt] Petersen, L.C.: On the relation between the multidimensional moment problem and the

one-dimensional moment problem, Math. Scand. 51(1982), 361–366.[PO] Petrovskii, I.G. and O.A. Oleinik: On the topology of real algebraic surfaces, Isvestiya

Akad. Nauk SSSR Ser. Mat. 13(1949), 389–402; Amer. Math. Translation 1992, No. 70.[Pi] Pick, G.: Über die Beschränkungen analytischer Funktionen, welche durch vorgegebene

Funktionswerte bewirkt sind, Math. Ann. 77(1916), 7–23.[Pl] Plaumann, D.: Sums of squares on reducible real curves, Math. Z. 265(2009), 777–797.[P] Polya, G.: Über positive Darstellung von Polynomen, Vierteljschr. Naturforsch. Ges.

Zürich 73(1928), 141–145.[PSz] Polya, G. and G. Szegö: Problems and Theorems in Analysis. II, Springer-Verlag, New

York, 1976.[PR] Powers, V. and B. Reznick: Polynomials that are positive on an interval, Trans. Amer.

Math. Soc. 352(2000), 4677–4692.[PoS] Powers, V. and C. Scheiderer: The moment problem for non-compact semi-algebraic

sets, Adv. Geom.1(2001), 71–88.[PD] Prestel, A. and C.N. Delzell: Positive Polynomials–from Hilbert’s 17th Problem to Real

Algebra, Springer-Verlag, New York, 2001.[Pu1] Putinar, M.: The L-moment problem in two dimensions, J. Funct. Analysis 94(1990),

288–307.[Pu2] Putinar, M.: Positive polynomials on compact semi-algebraic sets, Indiana Univ. Math.

J. 42(1994), 969–984.[Pu3] Putinar, M.: Extremal solutions of the two-dimensional L-problem of moments, J. Funct.

Analysis 136(1996), 331–364.[Pu4] Putinar, M.: A dilation approach to cubature formulas, Expo. Math 15(1997), 183–192.[Pu5] Putinar, M.: Extremal solutions of the two-dimensional L-problem of moments. II, J.

Approx. Theory 92(1998), 38–58.

524 Bibliography

[PSr] Putinar, V. and C. Scheiderer: Multivariate moment problems: geometry and indetermi-nateness, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (5), V(2006), 1–21.

[PSm] Putinar, N. and K. Schmüdgen: Multivariate determinateness, Indiana Univ. Math. J.57(2008), 2931–2968.

[PVs1] Putinar, N. and F.-H. Vasilescu: Solving the moment problem by dimension extension,Ann. Math. 149(1999), 1087–1107.

[PVs2] Putinar, N. and F.-H. Vasilescu: A uniqueness criterion in the multivariate momentproblem, Math. Scand. 92(2003), 295–300.

[RW] Rade, L. and B. Westergren: Springers Mathematische Formeln, Springer-Verlag, Berlin,1991.

[RS2] Reed, M. and B. Simon: Methods of Modern Mathematical Physics II. Fourier Analysisand Self-Adjointness, Academic Press, New York, 1975.

[RRS] Ren, Q., J. Richter-Gebert and B. Sturmfels: Cayley–Bacharach formulas, Amer. Math.Monthly 122(2015), 845–854.

[Re1] Reznick, B.: Sums of Even Powers of Linear Forms, Memoirs Amer. Math. Soc. 9,Providence, R.I., 1982.

[Re2] Reznick, B.: Uniform denominators in Hilbert’s seventheenth problem, Math. Z.220(1995), 75–98.

[Re3] Reznick, B.: Homogeneous polynomial solutions to constant PDE’s, Adv. Math.117(1996), 179–192.

[Re4] Reznick, B.: On Hilbert’s construction of positive polynomials, arXiv:0707.2156.[RiS] C. Riener and M. Schweighofer: Optimization approaches to quadrature: new charac-

terizations of Gaussian quadrature on the line and quadrature with few nodes on planealgebraic curves, on the plane and in higher dimensions, arXiv:1607.08404.

[Rz1] Riesz, F.: Über ein Problem des Herrn Carathéodory, J. reine angew. Math. 146(1916),83–87.

[Rz2] Riesz, M.: Sur le problème des moments. Premiere Note. Arkiv för Mat., Astr. och Fysik16(12) (1921), 23pp.; Deuxieme Note. Arkiv för Mat., Astr. och Fysik 16(19) (1922),21pp.; Troisieme Note. Arkiv för Mat., Astr. och Fysik 17(12) (1923), 52p.

[RzSz] Riesz, F. and B. Sz.-Nagy: Vorlesungen über Funltionalanalysis, DVW, Berlin, 1956.[Ri] Richter, H.: Parameterfreie Abschätzung und Realisierung von Erwartungswerten, Bl.

der Deutsch. Ges. Versicherungsmath. 3(1957), 147–161.[Rb] Robinson, R.M.: Some definite polynomials which are not sums of squares of real

polynomials, in: Selected Questions of Algebra and Logic, Acad. Sci. USSR, 1973,264–282.

[Rf] Rockafellar, R.T.: Convex Analysis, Princeton Univ. Press, Princeton, 1970.[Rg] Rogosinski, W.W.: Moments of non-negative mass, Proc. Roy. Soc. London Ser A

245(1958), 1–27.[Ru1] Rudin, W.: Functional Analysis, Second Edition, McGraw–Hill Inc., New York, 1973.[Ru2] Rudin, W.: Real and Complex Analysis, Third Edition, McGraw–Hill Inc., New York,

1987.[Sr1] Scheiderer, C.: Non-existence of degree bounds for weighted sums of squares represen-

tations, J. Complexity 2(2005), 823–844.[Sr2] Scheiderer, C.: Sums of squares on real algebraic curves. Math. Z. 245(2003), 725–760.[Sr3] Scheiderer, C.: Positivity and sums of squares: A guide to recent results, In: Emerging

Applications of Algebraic Geometry, M. Putinar and S. Sullivant (Editors), Springer-Verlag, Berlin, pp. 1–54, 2009.

[Sm1] Schmüdgen, K.: Positive cones in enveloping algebras, Reports Math. Phys. 14(1978),385–404.

[Sm2] Schmüdgen, K.: A positive polynomial which is not a sum of squares. A positive, butnot strongly positive functional, Math. Nachr. 88(1979), 385–390.

[Sm3] Schmüdgen, K.: A formally normal operator having no normal extension, Proc. Amer.Math. Soc. 98(1985), 503–504.

Bibliography 525

[Sm4] Schmüdgen, K.:Unbounded Operator Algebras and Representation Theory, Birkhäuser-Verlag, Basel, 1990.

[Sm5] Schmüdgen, K.: On determinacy notions for the two dimensional moment problem, Ark.Math. 29(1991), 277–284.

[Sm6] Schmüdgen, K.: The K-moment problem for compact semi-algebraic sets, Math. Ann.289(1991), 203–206.

[Sm7] Schmüdgen, K.: On the moment problem of closed semi-algebraic sets, J. reine angew.Math. 558(2003), 225–234.

[Sm8] Schmüdgen, K.: Noncommutative real algebraic geometry–some basic concepts and firstideas, In: Emerging Applications of Algebraic Geometry, M. Putinar and S. Sullivant(Editors), Springer, New York, 2009, 325–350.

[Sm9] Schmüdgen, K.: Unbounded Self-adjoint Operators on Hilbert Space, Springer-Verlag,New York, 2012.

[Sm10] Schmüdgen, K.: On the multi-dimensional truncated moment problem: maximal masses,Methods Funct. Anal. Topology 21(2015), 266–281.

[Sm11] Schmüdgen, K.: A fibre theorem for moments problems and some applications, Israel J.Math. 28(2016), 43–66.

[Sn] Schneider, R.: Convex Bodies: The Brunn–Minkowski Theory, Cambridge Univ. Press,Cambridge, 1993.

[SSz] Schoenberg, I.J. and G. Szegö: An extremum problem for polynomials, CompositioMath. 14(1960), 260–268.

[Sw1] Schweighofer, M.: An algorithmic approach to Schmüdgen’s Positivstellensatz, J. PureApplied Algebra 166(2002), 307–319.

[Sw2] Schweighofer, M.: Iterated rings of bounded elements and generalizations of Schmüd-gen’s Positivstellensatz, J. reine angew. Math. 554(2003), 19–45.

[Sw3] Schweighofer, M.: Optimization of polynomials on compact semi-algebraic sets, SIAMJ. Optimization 15(2005), 805–825.

[Su] Schur, I.: Über Potenzreihen, die im Innern des Einheitskreises beschränkt sind, I. J.reine angew. Math. 147(1917), 205–232, and II. 148(1918), 122–145.

[Sh] Shmuljan, J.L.: An operator Hellinger integral, Mat. Sb. 91(1959), 381–430.[SR] Semple, J.G. and L. Roth: Introduction to Algebraic Geometry, Clarendon Press, Oxford,

1949.[ST] Shohat, J.A. and J.D. Tamarkin: The Problem of Moments, Amer. Math. Soc., Provi-

dence, R.I., 1943.[Sim1] Simon, B.: The classical moment problem as a self-adjoint finite difference operator,

Adv. Math. 137 (1998), 82–203.[Sim2] Simon, B.: Orthogonal Polynomials on the Unit Circle, Part I: Classical Theory, Amer.

Math. Soc., Providence, R.I., 2005.[Sim3] Simon, B.: Szegö’s Theorem and Its Descendants, Princeton Univ. Press, Princeton,

2011.[Ste1] Stengle, G.: A Nullstellensatz and a Positivstellensatz in semialgebraic geometry, Math.

Ann. 207(1974), 87–97.[Ste2] Stengle, G.: Complexity estimates of Schmüdgen’s Positivstellensatz, J. Complexity 12

(1996), 167–174.[Stj] Stieltjes, T.J.: Recherches sur les fractions continues, Ann. Fac. Sci. Toulouse 8(1894),

1–122, and 9(1895), 1–47.[St1] Stochel, J.: Moment functions on real algebraic sets, Ark. Mat. 30(1992), 133–148.[St2] Stochel, J.: Solving the truncated moment problem solves the moment problem, Glasgow

J. Math. 43(2001), 335–341.[StS1] Stochel, J. and F.H. Szafraniec: On normal extrensions of unbounded operators. I, J.

Operator Theory 14(1985), 31–45.[StS2] Stochel, J. and F.H. Szafraniec: On normal extrensions of unbounded operators. II, Acta

Sci. Math. (Szeged) 53(1989), 153–177.

526 Bibliography

[StS3] Stochel, J. and F.H. Szafraniec: On normal extrensions of unbounded operators. III, Publ.RIMS Kyoto Univ. 25(1989), 105–139.

[StS4] Stochel, J. and F.H. Szafraniec: The complex moment problem and subnormality: a polardecomposition approach. J. Funct. Anal. 159(1998), 432–491.

[Stn] Stone, M.H.: A theory of spectra I, Proc. Nat. Acad. Sci. USA 26 (1940), 280–283.[Stv] Stoyanov, J.: Krein’s condition in probabilistic moment problems, Bernoulli 6(2000),

939–949.[SKv] Stoyanov, J. and P. Kopanov: Lin’s condition and moment determinacy of functions of

random variables, Preprint, 2017.[Sf] Szafraniec, F.H.: Boundedness of the shift operator related to positive definite forms: An

appplication to moment problems, Ark. Mat. 19(1981), 251–259.[Sz] Szegö, G.: Orthogonal Polynomials, Amer. Math. Soc., Providence, R.I., 1939.[SK] Sz.-Nagy, B. and A. Koranyi: Relations d’un probleme de Nevanlinna et Pick avec

la theorie des operateurs l’espace hilbertien, Acta Math. Acad. Sci. Hungar. 7(1956),295–303.

[Tch] Tchakaloff, V.: Formules de cubatures mécanique à coefficients non négatifs, Bull. Sci.Math. 82(1957), 123–134.

[To] Toeplitz, O.: Über die Fouriersche Entwicklung positiver Funktionen, Rend. Circ. Mat.Palermo 32(1911), 191–192.

[VA] Van Assche, V.: Orthogonal polynomials, associated polynomials and functions of thesecond kind, J. Comp. Appl. Math. 37(1991), 237–249.

[V] Vandenberghe, L. (Editor): Handbook of Semidefinite Programming: Theory, Algorithm,and Applications, Kluwer Academic, Boston, 2005.

[VB] Vandenberghe, L. and S. Boyd: Semidefinite programming, SIAM Rev. 38(1996), 49–95.

[Vs1] Vasilescu, F.: Hamburger and Stieltjes moment problems in several variables, Trans.Amer. Math. Soc. 354(2001), 1265–1278.

[Vs2] Vasilescu, F.: Spectral measures and moment problems, Theta Ser. Adv. Math. 2,Bucharest, 2003.

[Vs3] Vasilescu, F.: Dimensional stability in truncated moment problems, J. Math. Anal. Appl.388(2012), 219–230.

[Veg] Vegter, G.: The apolar bilinear form in geometric modeling, Math. Comp. 69(1999),691–720.

[Ver] Verblunsky, S.: On positive harmonic functions: A contribution to the algebra of Fourierseries, Proc. London Math. Soc. 38(1935), 125–157.

[Wr] Walker, R.J.: Algebraic curves, Springer-Verlag, New York, 1978.[Wl] Wall, H.S.: Analytic Theory of Continued Fractions, Amer. Math. Soc., Providence, R.I.,

2000.[Wei] Weidmann, J.: Linear Operators in Hilbert Spaces, Springer-Verlag, Berlin, 1987.[Wen] Wendroff, B.: On orthogonal polynomials, Proc. Amer. Math. Soc. 12(1961), 554–555.[Wö] Wörmann, T.: Strict positive Polynome in der semialgebraischen Geometrie, Disserta-

tion, Universität Dortmund, 1998.[Y] Young, N.: An Introduction to Hilbert Space, Cambridge Univ. Press, Cambridge, 1988.[Zh] Zhang, F.: Matrix Theory, Universitext, Springer-Verlag, Berlin, 1999.

Symbol Index

hA;Bi, 399, 505A;AT , 504AŒn�F ; A

Œn�K , 184

A 0;A 0, 504an, 98An.x; z/, 109A.y/, 399A.z;w/;A.z/, 147A, 18, 43, 415AC, 48AD, 332Ah, 43OA, 18A2, 44A2, 425˛.n/, 489˛n.s/; ˛n.�/, 264

B.K.f//, 293bn, 98Bn.x; z/, 109B.z;w/;B.z/, 147B.H/, 514B.X /, 499ˇs, 191

Cc.X /;Cc.X IR/;Cc.X /C;C0.X /, 499Cf , 344cL.x/, 32, 457Cfmng, 81Cn.x; z/, 109C^, 511C.X IR/, 13

Cz, 151C.z;w/;C.z/, 147Cd;m, 482cmC1, 231, 266C.s/;C.L/;C.A;K/, 449CC, 502

Dm.s/;Dm.s/, 230Dn;Dn.s/, 93Dn.s/, 63Dn.x; z/, 109dC.T/; d�.T/, 515D.z;w/;D.z/, 147d, 123DL, 303, 425DŒs�, 177D.T/, 514D, 271D.Rd Œ x �/, 332

EC, 14Es, 63E�, 511ET , 516eEt;Et, 464Exr.C/, 513�.t/, 232

fs, 218Ef , 230, 426fz.x/, 125f�, 322


527

528 Symbol Index

�s, 191�n. f /; �n.�/, 274

hd;2n, 480H.gs/, 290H.L/, 425H2n.s/;H2n.s/;H2nC1.s/;H2nC1.s/, 230Hn.s/, 63H.'/, 45Hz;w;Hz, 152HdC1;2n, 420Hd;m, 471Hs, 102Hd

1, 420

I�.z/, 503It. f ; g/, 488I�; OI�, 322

J; TJ , 103

KStz , 198

Kz; z;Cz, 151K.f/;K. f1; : : : ; fk/, 284K.T/�; T�; f�; I�; OI�, 322K.Q/, 288K.Q/�, 323K, 43KŒS�, 45 L.x/, 26, 457 s.�/, 241

L.A;K/, 446Lg, 289L�, 14L' , 45lx, 22L, 24L.K/, 19L1.X ; �/, 500

Mn;m.K/;Mn.K/, 504Mn.z/;Nn.z/, 185MC.X /;M1

C.X /;Mb

C.X /;M.X /, 499

ML;K ;ML, 14, 416MC.OA/, 320MC.R/, 82MC.R

d/, 290

Ms, 36ind.�/, 232�A, 122�F; �K , 180�C; ��, 238�t, 149�� , 240

Nk.L/, 34Nn.z/, 185NC.L/, 30Nd;m, 471N2d0 , 51

Nk.L/, 453NL, 303, 425NC.L;K/;NC.L/, 451NC.L/, 292Ns, 217N .T/, 514Nd0 , 49

�sym, 67

ord.�/, 165

prz.x/; pnrz.x/, 69Op, 491Pk.x/;Pk.x/, 248pk.x/; pk.x/, 249

pmin, 403pmomn , 403Pn.x/, 96pn.x/, 95Pos.K/, 18Pos.M/n, 58Pos.A;K/, 418Pos.K/2n , 419Œ p; q�, 472h p; qis, 94, 122@˛; p.@/, 475psosn , 403.PSP/A;K, 461Pd;2n, 477P.f/, 307P.N0/, 63Pd.R/, 420P, 502pz, 105�j; �j.�/, 362�L, 303, 433

Symbol Index 529

Q.f/;Q. f1; : : : ; fk/, 285qh, 491Q

k.x/;Qk.x/, 248

qk.x/; qk.x/, 249

Q�, 323Qn.z/, 107qn.z/, 105Qsat, 321Qd;m, 477qz, 105

RŒV�, 286rk.s/, 218L.x/, 456.T/, 514z, 151s.�/, 241

s˛.�/, 1, 50.S; ı/; .S; ı;�/, 45Sd�1, 298Sd�1

C, 336

sŒ j�n , 368.SP/A , 451sC; s�, 238�, 43P

A2, 44Symn, 399S , 36S.A;K/, 446SmC1, 231, 266�.T/, 514s.x/, 29sN.x/; s.x/, 429�, 284P2

d;2n, 477supp �, 500P.WI g1; : : : ; gk/, 339

TF ; TK , 178T.f/; T. f1; : : : ; fk/, 285T; TJ , 103T�, 322Œt0 W � � � W td�, 420T; T�, 515Tt; T1, 134T , 108�f , 509�L.t/, 465TrA, 505

Vj.L/;V.L/, 34VC.L/, 30Vj.L/, 453V.L/, 453VL, 425VC.L;K/;VC.L/, 451

w. f /;w.L/, 482W.�; ˇ/n, 108W.L/, 28W.z; t/, 196W.L/, 454W.L;K/, 456

X, 102X , 13

.y�/m; .y � x/m , 472

ZD , 332Z. f /, 22ZP. f /, 488Z.S/, 285Zd , 52

Index

Adapted space, 15�-Algebra, 43

complexification, 48Hermitian part, 43involution, 43

Apolar scalar product, see Homogeneouspolynomials

Basic closed semi-algebraic set, 285Blaschke product, 273Block matrix

flat extension, 506positive semidefinite, 506

Borel algebra, 499Borel function, 499

Carathéodory number, 449Carleman condition, 80, 90

complex, 391multivariate, 369

Catalecticant matrix, 51Cauchy–Schwarz inequality, 44Character, 18, 46, 47Christoffel–Darboux formula, 109Complex moment problem, 382

for the d-torus Td , 385for the cylinder set R � Td , 385operator-theoretic approach, 389two-sided, 395

Cone, 284, 511base, 513dual, 511exposed face, 513

extreme ray, 513face, 513pointed, 513supporting functional, 511

Conic optimization problem, 514Continued fraction, 140Convex set, 511

extreme point, 513relatively interior point, 511separation theorem, 509

Core variety, 34, 453Cramer condition, 85

Deficiency indices, 515Disintegration theorem for measures, 374

Entropy integral, 87

Farkas lemma, 307Fibre theorem

for determinacy, 375for preorderings, 323, 349, 352for quadratic modules, 323, 349

Finest locally convex topology, 321Friedrichs extension, 178Frobenius lemma, 218Function

K-moment, 48Carathéodory, 271completely monotone, 66lower semicontinuous, 501positive semidefinite, 45quasi-analytic, 81


531

532 Index

Schur, 271upper semicontinuous, 501

FunctionalK-moment, 14, 47

K-determinate, 14, 47localized, localization, 289positive, 44Riesz, 45strictly EC-positive, 26

Gaussian quadrature formula, 207Gelfand–Naimark–Segal construction,

GNS-construction, 303Generalized Hankel matrix, 46Goursat transform, 61Gram matrix, 317

Hamburger moment problem, 63determinate, 80, 126existence of a solution, 63, 121, 210indeterminate, 86, 128, 130, 145

N-extremal, 150finite order solutions, 165Nevanlinna parametrization, 155von Neumann solution, 158Weyl circles, 151

strong, 66symmetric, 67, 136two-sided, 66von Neumann solution, 123, 128, 149, 151

Hankel determinant, 63, 93, 230Hankel matrix, 50, 63, 229, 425

generalized, 46localized, 290

Hardy condition, 85Hausdorff moment problem, 66Heine formulas, 96Homogeneous polynomials

apolar scalar product, 472apolarity lemma, 476basic set of nodes, 473differential operators, 476Marsden identity, 474

Carathéodory number, 482dehomogenization, 491truncated moment functional, 479

Jacobi matrix, 103Jacobi operator, 103, 123

self-adjoint extensions, 134

Krein condition, 85Krein extension, 178

Lasserre relaxations, 403Linear matrix inequality, 400Locally convex topology, 509

finest, 509Logarithmic integral, 87Lognormal distribution, 89

Marginal measure, 362Matrix

Hermitian, 504positive definite, 504positive semidefinite, 504spectral theorem, 504

Maximal mass, 241, 457Measure

Radon, 499representing, 14

Moment cone, 231, 266, 446Moment curve, 231, 266Moment function, 48

determinate, 48Moment functional, 14, 47

determinate, 14, 47extreme point of representing measures, 21width, 482with bounded density, 297, 379

Moment generating function, 91Moment matrix, 51Moment problem, 1, 13

for d-dimensional compact intervals, 309for compact intervals, 65for compact semi-algebraic sets, 295, 296,

302for cubics, 345for cylinder sets with compact base, 330for sets contained in compact polyhedra,

308for the half-line, see Stieltjes moment

problemfor the real line, see Hamburger moment

problemfor unit spheres, 298

Moment property (MP), 320Moment sequence, 50

Hamburgerindeterminate, maximal mass, 157

Momentsof a measure, 1of the Gaussian measure, 373

Index 533

of the surface measure of the unit sphere,373

Multidimensional determinacy notionsK-determinate, 363determinate, 358strictly determinate, 358strongly determinate, 358ultradeterminate, 358

Multidimensional Hausdorff moment problem,311

Multidimensional rational moment problem,333

Natural topology, 509Nevanlinna functions, 147Nevanlinna–Pick interpolation, 161

Operatoradjoint, 515closed, 514conjugation, 515domain, 514essentially self-adjoint, 515formally normal, 386kernel, 514normal, 386resolvent set, 514self-adjoint, 515

spectral theorem, 516spectrum, 514symmetric, 514

deficiency indices, 515Orthogonal polynomials, 96

Chebyshev, 117Hermite, 118Laguerre, 118Legendre, 118monic, 96of the first kind, 94reproducing kernel, 173zeros, 112of the second kind, 105

Orthonormal polynomials, 94

Pick function, 155, 160, 193, 502canonical representation, 502

PolynomialChoi–Lam, 353homogenization, 491Motzkin, 318Robinson, 483

Polynomial optimizationglobal, 408over a semi-algebraic set, 403

Portmanteau theorem, 501Positive functional, 44Positivstellensatz

Archimedean, 302Krivine–Stengle, 285strict, 295

Preordering, 284Preprime, 284

Quadratic module, 284Archimedean, 286, 305

representation theorem, 301closed, 342saturated, 288stable, 339

Quasi-analytic vector, 372Quasi-orthogonal polynomial, 204

Radon measure, 499complex, 500support, 500symmetric, 67

Rational multidimensional moment problem,333

Real algebraic set, 285algebra of regular functions, 286

Real ideal, 323Real projective curve, 488

cubic, 492quadratic, 492

Real projective space, 420Real radical, 323Real Waring rank, 474�-Representation, 302

algebraically cyclic vector, 302domain, 302

Representing measure, 14Reproducing kernel, 173, 212Riesz functional, 45Riesz’ representation theorem, 500

Schur algorithm, 274Schur parameters, 274Semi-algebraic set, 285Semidefinite program

dual program, 400duality gap, 401feasible vector, 400

534 Index

primal program, 399strictly feasible vector, 401

Semigroup, 45�-Semigroup, 45

character, 47Semigroup �-algebra, 45Semiring, 284

Archimedean, 286representation theorem, 301

Sequencecomplex moment, 382moment, 50positive definite, 64positive semidefinite, 50, 63

Set of atoms, 28, 29, 34Sos representation, 317Spectrahedron, 400Spectral theorem

for self-adjoint operators, 122, 179, 516for strongly commuting normal operators,

387for strongly commuting self-adjoint

operators, 304, 360Stieltjes moment problem, 65

determinate, 81, 182existence of a solution, 65, 179Friedrichs approximant, 184Friedrichs solution, 180indeterminate

Friedrichs parameter, 193Nevanlinna parametrization, 193Weyl circles, 198

Krein approximant, 184Krein solution, 180two-sided, 76

Stieltjes transform, 503Stieltjes–Perron inversion formula, 503Strong moment property (SMP), 320Support of a Radon measure, 500

TheoremBernstein, 60Bezout, 489Biermann, 473Bisgaard, 396Carathéodory, 512Carathéodory–Toeplitz, 260Carleman, 80, 132Chasles, 494Curto–Fialkow, 436Denjoy–Carleman, 81Eidelheit, 509Favard, 100, 104

Fejér–Riesz, 257Geronimus, 277Hamburger, 63, 121, 210Haviland, 17, 18Herglotz, 271Hilbert, 480, 483Krein, 85Krivine–Stengle, 285Lebesgue, 500Markov, 139, 210Markov–Lukacs, 74Nevanlinna, 155Petersen, 362Polya, 61, 310Radon–Nikodym, 500Reznick, 312Richter–Tchakaloff, 23Riesz, 500Schur, 53Shmuljan, 506Stieltjes, 65Stochel, 419Stochel–Szafraniec, 393Verblunsky, 264

Three term recurrence relation, 99, 100, 104Toeplitz matrix, 52, 260, 266Trigonometric moment problem, 257

existence of a solution, 260for circular arcs, 279orthogonal polynomials, 262reflection coefficients, 264, 271Schur parameters, 264Szegö recursion formulas, 263Verblunsky coefficients, 264

Truncated Hamburger moment problem, 203Christoffel function, 212existence of a solution, 209, 216, 221,

223Hankel rank, 218minimal polynomial, 218reproducing kernel, 212

Truncated moment problem on a boundedinterval, 229

canonical measure, 235, 247, 252determinacy, 233existence of a solution, 230index of representing measure, 232maximal mass, 241moment cone, 231

boundary point, 233interior point, 235

moment curve, 231orthogonal polynomials, 248principal measure, 237, 252

Index 535

Truncated multidimensional moment problem,415

Carathéodory number, 449core variety, 34, 453determinate moment functionals, 29, 455evaluation polynomials, 464

orthogonal family, 466extreme points of representing measures,

450flat extension of a functional, 435flat extension theorem, 435Hankel matrix, 425

finitely atomic measure, 429homogeneous polynomials, 479maximal mass, 457, 459maximal mass measure, 459minimal functional, 431moment cone, 446necessary conditions for truncated moment

functional, 430on projective space, 421

ordered maximal mass measure, 462positive separation property, 461separation property, 451set of atoms, 28, 34, 454truncated K-moment sequence, 416

Truncated Stieltjes moment problem, 203existence of a solution, 225, 226

Truncated trigonometric moment problem, 266canonical measure, 268moment cone, 266moment curve, 266

Vague convergence, 501Vague topology, 500

Weyl circles, 151, 198Wronskian, 108

Plücker identity, 143

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